contestada

Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to occur, the ratio of the amounts of propane to oxygen must be adjusted correctly. a. Use the balanced equation C3H8 + 5O2 3CO2 + 4H2O to answer the following questions. (11 points)i. How many moles of CO2 are produced from 5 moles O2? (1 point) ii. How many grams of CO2 are produced from 5 moles O2? (2 points) iii. How many grams of CO2 are produced from 128.00 g O2? (2 points) iv. A total of 132.33 g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed. (6 points) • Determine if one of the reactants is a limiting reagent. (3 points)

Respuesta :

aldana20

Answer:

1- 3 Moles of CO2  

2- 132 g of CO2  

3- 105,6 g of CO2

4- Limiting Reagent O2

Products form based on limiting reagent (384g O2) :

CO2: 316,8 g

H2O: 172,8 g

Products form based on C3H8 (132,33 g):

CO2: 396,99 g

H2O: 216,54 g  

Explanation:

Atomic Masses:

C: 12

H: 1

O: 16

Molecular weights:

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g  O2 ⇒ 132 g CO2

128 g  O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For          44 g of CH3H8  ⇒160 g of O2

With   132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

Calculations of the products formed in base of quantity of O2 (limiting reagent):

160 g  O2 ⇒ 132 g CO2

384 g  O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g  O2 ⇒ 72 g H2O

384 g  O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

Calculations of the products formed in base of quantity of C3H8 (excess reagent):

     44 g  C3H8 ⇒ 132 g CO2

132,33 g  C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

     44 g C3H8 ⇒ 72 g H2O

132,33 g  C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

Answer:

a.

C3H8 + 5 O2 → 3 CO2 + 4 H2O

i.

(5 mol O2) x (3 mol CO2 / 5 mol O2) = 3 mol CO2

ii.

(5 mol O2) x (3 mol CO2 / 5 mol O2) x (44.00964 g CO2/mol) = 132 g CO2

iii.

(128.00 g O2) / (31.99886 g O2/mol) x (3 mol CO2 / 5 mol O2) x  

(44.00964 g CO2/mol) = 105.63 g CO2

iv.

(132.33 g C3H8) / (44.09592 g C3H8/mol) = 3.000958 mol C3H8

(384.00 g O2) / (31.99886 g O2/mol) = 12.00043 mol O2

12.00042 moles of O2 would react completely with 12.00042 x (1/5) = 2.400084 mol C3H8, but there is more C3H8 present than that, so C3H8 is in excess and O2 is the limiting reagent.

(12.00043 mol O2) x (3 mol CO2 / 5 mol O2) x (44.00964 g CO2/mol) =  

316.88 g CO2

(12.00043 mol O2) x (4 mol H2O / 5 mol O2) x (18.01532 g H2O/mol) =  

172.95 g H2O

b.  

(269.34 g CO2) / (316.88 g CO2) = 0.84997 = 84.997% yield

Explanation:

WANNA GIMME BRAINLY????

Otras preguntas