myabean365
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85 POINTS!!!
1.) 2HgO(s) --> 2Hg(l) + O2 (g)
In a certain reaction 8.74 g of HgO is decomposed, producing 6.42 g of Hg. What is the percent yield for the reaction?

2.) Fluorine gas reacts with a solution of sodium bromide in a single-replacement reaction producing aqueous sodium fluoride and bromine gas.
If a reaction starts with 0.240 mol fluorine gas, how many moles of sodium fluoride are produced?

3.) 4 Fe(s) + 3O2(g) -->2Fe2O3(g)
In a certain reaction, 54.6 g of iron is reacted with 91.6 g of Oxygen
How many grams of Iron (III) oxide are produced in the reaction?

4.) Ga2O3(s) + 3SOCl2(l) --> 2GaCl3(s) +3SO2
In a certain reaction, 71.8 g of Ga2O3 is reacted with 110.8 g SOCl2.The GaCl3 produced is collected and its mass founded to be 97.66 g.
What is the theoretical yield of GaCl3?

5.) N2 (g) + 3H2 (g) -->2 NH3 (g)
The equation above is the equation for the Haber process.
In a certain reaction, you start with 6.0 moles of nitrogen and 10.0 moles of hydrogen,
which is the excess reactant?

6.) B2O3(s) + 3C(s) + 3Cl2 (g) ---> 2 BCl3(g) + 3CO(g)
What is the mass of carbon required in order to produce 3.00 mol BCl3?

7.) B2O3(s) + 3C(s) + 3Cl2 (g) ---> 2 BCl3(g) + 3CO(g)
Calculate how many moles of BCl3 are produced from the reaction of 122.0g of B203?

8.) 4 Fe(s) + 3O2(g) -->2Fe2O3(g)
In a certain reaction, 54.6 g of iron is reacted with 91.6 g of Oxygen
How many grams of the excess reactant remain after the reaction is complete?

9.) 2NO(g) + O2 (g) --> 2NO2
How many grams of NO are required to react fully with 2.50 mol of O2?

10.) N2 (g) + 3H2 (g) -->2 NH3 (g)
The equation above is the equation for the Haber process.
In a certain reaction, you start with 3.0 moles of nitrogen and 5.0 moles of hydrogen, which molecule is the limiting reagent?

11.) 2HgO(s) --> 2Hg(l) + O2 (g)
In a certain reaction 8.74 g of HgO is decomposed, producing 6.42 g of Hg.
What is the theoretical yield of Hg?

12.) B2O3(s) + 3C(s) + 3Cl2 (g) ---> 2 BCl3(g) + 3CO(g)
How many moles of chlorine are used up in a reaction that produced 175 g of BCl3?

13.) 4 Fe(s) + 3O2(g) --> 2Fe2O3(g)
In a certain reaction, 54.6 g of iron is reacted with 91.6 g of Oxygen
What is the limiting reactant?

14.) N2 (g) + 3H2 (g) -->2 NH3 (g)
The equation above is the equation for the Haber process.
In a certain reaction, you start with 6.0 moles of nitrogen and 10.0 moles of hydrogen. How many moles of ammonia will be produced in the reaction?

15.) B2O3(s) + 3C(s) + 3Cl2 (g) ---> 2 BCl3(g) + 3CO(g)
How many moles of carbon are used up in a reaction that produced 0.70 kg of BCl3?

16.) Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g)
A certain reaction produced 7.40 mol carbon dioxide. How many moles of both Fe2O3 and CO reacted?

Respuesta :

ChemistryHelper2024
1)    2 HgO    →      2 Hg + O₂
  (2 x 216.59 g) → gives → (2 x 200.59 g)
   433.18 g   → gives → 401.18 g
     8.74 g     → gives → ? g
by cross multiplication:
Theoretical yield = (401.18 x 8.74) / 433.18 = 8.094 g
% yield = (actual yield / theoretical yield) * 100
             = (6.42 / 8.09) * 100 = 79.3 % 

2) F₂ + 2 NaBr → 2 NaF + Br₂
1 mole F₂ → gives → 2 moles NaF
0.24 mol F₂ will give ?? mol NaF
= (0.24 x 2) / 1 = 0.48 mol NaF

3) 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)
number of moles of Fe = (54.6 / 55.845) = 0.97 mol
number of moles of O₂ = (91.6 g / 32) = 2.8625 mol
4 moles Fe react with 3 moles O₂
0.97 mol Fe will react completely with 0.7275 mol O₂
so 0.97 mol Fe is the limiting reactant and O₂ present in excess
4 moles Fe(s)    → gives → 2 moles Fe₂O₃  
0.97 mol Fe(s)  → gives → ? moles Fe₂O₃
= (0.97 x 2) / 4 = 0.485 mol Fe₂O₃
mass of Fe₂O₃ = 0.485 mol * 159.69 g/mol = 77.45 g

4) Ga₂O₃(s) + 3 SOCl₂(l) → 2 GaCl₃(s) + 3 SO₂
Molar mass of Ga₂O₃ = 187.44 g/mol
Molar mass of SOCl₂ = 118.97 g/mol
Mass of Ga₂O₃ = 71.8 g 
Mass of SOCl₂ = 110.8 g 
number of moles = mass (g) / molar mass 
number of moles of Ga₂O₃ = 71.8 / 187.44 = 0.38 mol
number of moles of SOCl₂ = 110.8 / 118.97 = 0.93 mol
1 mole Ga₂O₃(s) react with 3 moles SOCl₂
0.38 mol Ga₂O₃ react with 1.14 mole SOCl₂
But only 0.93 mol SOCl₂ present so it is the limiting reactant
3 moles of SOCl₂ → gives → 2 moles GaCl₃
0.93 mol SOCl₂  → gives → ? mole GaCl₃
= (2 x 0.93) / 3 = 0.62 mole
Mass of GaCl₃ = 0.62 * 176.073 = 109.16 g

5) N₂(g) + 3 H₂(g) → 2 NH₃(g)
   1 mol N₂ reacts with 3 moles H₂
 so 6 moles N₂ will react completely with 18 moles H₂
But moles of H₂ present is only 10 moles, so Nitrogen is considered as the excess reactant.

6) B₂O₃(s) + 3 C(s) + 3 Cl₂(g)   → 2 BCl₃(g) + 3 CO(g)
3 moles C → gives → 2 moles BCl₃
? moles C → gives → 3 moles BCl₃
 moles of C = (3*3) / 2 = 4.5 moles C
Mass of carbon required = 4.5 moles * 12 g/mol = 54 g carbon

7) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2 BCl₃(g) + 3 CO(g)
number of moles of B₂O₃ =  mass / molar mass
                                       = 122 g / 69.6182 = 1.75 moles 
1 mole B₂O₃       → gives → 2 moles BCl₃
1.75 mole B₂O₃  → gives → ? moles BCl₃
= 2 * 1.75 = 3.5 moles BCl₃ 

8) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃
moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol
moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol
4 moles Fe(s) react with 3 moles O₂(g)
0.98 mol Fe(s) react completely with ? moles O₂(g)
= 0.735 mol O₂
moles of excess oxygen = 2.8625 - 0.735 = 2.1275 mol 
mass of excess oxygen = 2.1275 * 32 = 68.08 g O₂

9) 2 NO(g) + O₂(g) → 2 NO₂ 
2 moles NO react with 1 mol O₂
? moles NO will react with 2.5 mol of O₂
= (2 * 2.5) / 1 = 5 moles NO
Mass of NO = 5 * 30 = 150 g

10) N₂(g) + 3 H₂(g) → 2 NH₃(g)
1 mole N₂ reacts with 3 moles H₂
so 3 moles of N₂ will react completely with 9 moles H₂
but only 5 moles of Hydrogen present so it is the limiting reactant

11) 2 HgO(s)        →      2 Hg(l) + O₂(g)
(2 x 216.59 g) give    (2 x 200.59)
433.18 g will give    401.18 g
8.74 g HgO  will give ? g Hg
Theoretical yield of Hg = (8.74 x 401.18) / 433.18 = 8.094 g

12) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)
number of moles of BCl₃ produced = 175 g / 117.17 g/mol = 1.5 moles
3 moles of Cl₂ produce 2 moles BCl₃
? moles of Cl₂ produces 1.5 mole BCl₃
= (1.5 x 3) / 2 = 2.25 moles Cl₂

13) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃
moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol
moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol
4 moles Fe(s) react with 3 moles O₂(g)
0.98 mol Fe(s) react completely with ? moles O₂(g)
= 0.735 mol O₂
So the limiting reactant is Fe

14) N₂(g) + 3 H₂(g) → 2 NH₃(g)
1 mole N₂ reacts with 3 moles H₂
so 6 moles of N₂ will react completely with 18 moles H₂
but only 10 moles of Hydrogen present so it is the limiting reactant
3 moles of H₂ gives 2 moles NH₃
10 moles of H₂ will give ? moles NH₃
= (10 * 2) / 3 = 6.67 moles NH₃

15) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)
number of moles of BCl₃ produced = 700 g / 117.17 g/mol = 5.97 moles
3 moles of C produce 2 moles BCl₃
? moles of C produces 5.97 mole BCl₃
= (5.97 x 3) / 2 = 8.96 moles C

16) Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3CO₂(g)
1 mole of Fe₂O₃ gives 3 moles CO₂
? mole of Fe₂O₃ gives 7.4 moles CO₂
= 2.47 moles Fe₂O₃
3 moles of CO gives 3 moles of CO₂
so 7.4 moles of CO will produce the required amount of CO₂ (7.4)






adamaek00

1)    2 HgO    →      2 Hg + O₂

 (2 x 216.59 g) → gives → (2 x 200.59 g)

  433.18 g   → gives → 401.18 g

    8.74 g     → gives → ? g

by cross multiplication:

Theoretical yield = (401.18 x 8.74) / 433.18 = 8.094 g

% yield = (actual yield / theoretical yield) * 100

            = (6.42 / 8.09) * 100 = 79.3 % 

2) F₂ + 2 NaBr → 2 NaF + Br₂

1 mole F₂ → gives → 2 moles NaF

0.24 mol F₂ will give ?? mol NaF

= (0.24 x 2) / 1 = 0.48 mol NaF

3) 4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)

number of moles of Fe = (54.6 / 55.845) = 0.97 mol

number of moles of O₂ = (91.6 g / 32) = 2.8625 mol

4 moles Fe react with 3 moles O₂

0.97 mol Fe will react completely with 0.7275 mol O₂

so 0.97 mol Fe is the limiting reactant and O₂ present in excess

4 moles Fe(s)    → gives → 2 moles Fe₂O₃  

0.97 mol Fe(s)  → gives → ? moles Fe₂O₃

= (0.97 x 2) / 4 = 0.485 mol Fe₂O₃

mass of Fe₂O₃ = 0.485 mol * 159.69 g/mol = 77.45 g

4) Ga₂O₃(s) + 3 SOCl₂(l) → 2 GaCl₃(s) + 3 SO₂

Molar mass of Ga₂O₃ = 187.44 g/mol

Molar mass of SOCl₂ = 118.97 g/mol

Mass of Ga₂O₃ = 71.8 g 

Mass of SOCl₂ = 110.8 g 

number of moles = mass (g) / molar mass 

number of moles of Ga₂O₃ = 71.8 / 187.44 = 0.38 mol

number of moles of SOCl₂ = 110.8 / 118.97 = 0.93 mol

1 mole Ga₂O₃(s) react with 3 moles SOCl₂

0.38 mol Ga₂O₃ react with 1.14 mole SOCl₂

But only 0.93 mol SOCl₂ present so it is the limiting reactant

3 moles of SOCl₂ → gives → 2 moles GaCl₃

0.93 mol SOCl₂  → gives → ? mole GaCl₃

= (2 x 0.93) / 3 = 0.62 mole

Mass of GaCl₃ = 0.62 * 176.073 = 109.16 g

5) N₂(g) + 3 H₂(g) → 2 NH₃(g)

  1 mol N₂ reacts with 3 moles H₂

 so 6 moles N₂ will react completely with 18 moles H₂

But moles of H₂ present is only 10 moles, so Nitrogen is considered as the excess reactant.

6) B₂O₃(s) + 3 C(s) + 3 Cl₂(g)   → 2 BCl₃(g) + 3 CO(g)

3 moles C → gives → 2 moles BCl₃

? moles C → gives → 3 moles BCl₃

 moles of C = (3*3) / 2 = 4.5 moles C

Mass of carbon required = 4.5 moles * 12 g/mol = 54 g carbon

7) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2 BCl₃(g) + 3 CO(g)

number of moles of B₂O₃ =  mass / molar mass

                                      = 122 g / 69.6182 = 1.75 moles 

1 mole B₂O₃       → gives → 2 moles BCl₃

1.75 mole B₂O₃  → gives → ? moles BCl₃

= 2 * 1.75 = 3.5 moles BCl₃ 

8) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃

moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol

moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol

4 moles Fe(s) react with 3 moles O₂(g)

0.98 mol Fe(s) react completely with ? moles O₂(g)

= 0.735 mol O₂

moles of excess oxygen = 2.8625 - 0.735 = 2.1275 mol 

mass of excess oxygen = 2.1275 * 32 = 68.08 g O₂

9) 2 NO(g) + O₂(g) → 2 NO₂ 

2 moles NO react with 1 mol O₂

? moles NO will react with 2.5 mol of O₂

= (2 * 2.5) / 1 = 5 moles NO

Mass of NO = 5 * 30 = 150 g

10) N₂(g) + 3 H₂(g) → 2 NH₃(g)

1 mole N₂ reacts with 3 moles H₂

so 3 moles of N₂ will react completely with 9 moles H₂

but only 5 moles of Hydrogen present so it is the limiting reactant

11) 2 HgO(s)        →      2 Hg(l) + O₂(g)

(2 x 216.59 g) give    (2 x 200.59)

433.18 g will give    401.18 g

8.74 g HgO  will give ? g Hg

Theoretical yield of Hg = (8.74 x 401.18) / 433.18 = 8.094 g

12) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)

number of moles of BCl₃ produced = 175 g / 117.17 g/mol = 1.5 moles

3 moles of Cl₂ produce 2 moles BCl₃

? moles of Cl₂ produces 1.5 mole BCl₃

= (1.5 x 3) / 2 = 2.25 moles Cl₂

13) 4 Fe(s) + 3 O₂ → 2 Fe₂O₃

moles of Fe = (54.6 g) / (55.845 g/mol) = 0.98 mol

moles of O₂ = (91.6 g) / (32 g/mol) = 2.8625 mol

4 moles Fe(s) react with 3 moles O₂(g)

0.98 mol Fe(s) react completely with ? moles O₂(g)

= 0.735 mol O₂

So the limiting reactant is Fe

14) N₂(g) + 3 H₂(g) → 2 NH₃(g)

1 mole N₂ reacts with 3 moles H₂

so 6 moles of N₂ will react completely with 18 moles H₂

but only 10 moles of Hydrogen present so it is the limiting reactant

3 moles of H₂ gives 2 moles NH₃

10 moles of H₂ will give ? moles NH₃

= (10 * 2) / 3 = 6.67 moles NH₃

15) B₂O₃(s) + 3 C(s) + 3 Cl₂(g) → 2BCl₃(g) + 3 CO(g)

number of moles of BCl₃ produced = 700 g / 117.17 g/mol = 5.97 moles

3 moles of C produce 2 moles BCl₃

? moles of C produces 5.97 mole BCl₃

= (5.97 x 3) / 2 = 8.96 moles C

16) Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3CO₂(g)

1 mole of Fe₂O₃ gives 3 moles CO₂

? mole of Fe₂O₃ gives 7.4 moles CO₂

= 2.47 moles Fe₂O₃

3 moles of CO gives 3 moles of CO₂

so 7.4 moles of CO will produce the required amount of CO₂ (7.4)

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