Respuesta :
Answer: [tex]CaCl_2[/tex] is the limiting reagent and theoretical yield of [tex]CaCO_3[/tex] is 8 grams.
Explanation:
[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]
Thus [tex]\text{no of moles}of Na_2CO_3={3.0M}\times {0.06 L}=0.18moles[/tex]
Thus [tex]\text{no of moles}of CaCl_2={2.0M}\times {0.04 L}=0.08moles[/tex]
[tex]Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl[/tex]
As 1 mole of [tex]Na_2CO_3[/tex] combines with 1 mole of [tex]CaCl_2[/tex]
Thus 0.08 moles of [tex]CaCl_2[/tex] react with =[tex]\frac{1}{1}\times 0.08=0.08[/tex] moles of [tex]Na_2CO_3[/tex]
Thus [tex]CaCl_2[/tex] is the limiting reagent as it limits the formation of products and [tex]Na_2CO_3[/tex] is an excess reagent.
1 moles of [tex]CaCl_2[/tex] form = 1 mole of [tex]CaCO_3[/tex]
0.08 moles of [tex]CaCl_2[/tex] form =[tex]\frac{1}{1}\times 0.08=0.08[/tex] moles of [tex]CaCO_3[/tex]
[tex]{\text {Mass of}CaCO_3=moles\times {\text {Molar Mass}}=0.08moles\times 100g/mol=8g[/tex]
Thus theoretical yield of [tex]CaCO_3[/tex] is 8 grams.
- CaCl₂ acts as the limiting reagent.
- The theoretical yield of CaCO₃ is 80 mmol or 0.08 moles.
Further explanation
Given:
- 60.0 mL of 3.00 M Na₂CO₃
- 40.0 mL of 2.00 M CaCl₂
Question:
- What is the limiting reagent?
- What is the theoretical yield of CaCO₃ in moles?
The Process:
Step-1: prepare moles for each reagent
[tex]\boxed{ \ Molarity = \frac{moles}{volume} \ }[/tex] [tex]\rightarrow \boxed{ \ n = MV \ }[/tex]
[tex]\boxed{ \ Na_2CO_3 \ }[/tex] [tex]\rightarrow \boxed{ \ n = 3 \ \frac{mol}{L} \times 60 \ mL \ }[/tex] [tex]\rightarrow \boxed{ \ n = 180 \ mmol \ }[/tex]
[tex]\boxed{ \ CaCl_2 \ }[/tex] [tex]\rightarrow \boxed{ \ n = 2 \ \frac{mol}{L} \times 40 \ mL \ }[/tex] [tex]\rightarrow \boxed{ \ n = 80 \ mmol \ }[/tex]
Step-2: the ICE table
We use the ICE table to see how the reaction occurs between two salts.
Balanced reaction:
[tex]\boxed{ \ Na_2CO_3 + CaCl_2 \rightarrow CaCO_3 + 2NaCl \ }[/tex]
Initial: 180 80 - -
Change: 80 80 80 80
Equlibrium: 100 - 80 80
- Notice that the coefficient ratio between Na₂CO₃ and CaCl₂ is 1 : 1. Remember, the coefficient ratio is also the mole ratio of the reactants.
- Based on this ratio, it appears that the initial amount of Na₂CO₃ is more than CaCl₂, or the amount of Na₂CO₃ is known to be excessive.
- CaCl₂ acts as the limiting reagent because the amount does not remain at the end of the reaction.
Final step: calculate the theoretical yield of CaCO₃ in moles and grams
From the ICE table above, the theoretical yield of CaCO₃ produced is 80 mmol or 0.08 moles.
Let us try to continue by calculating the mass of CaCO₃ produced.
- Relative atomic mass: Ca = 40, C = 12, and O = 16.
- Relative molecular mass (Mr) of CaCO₃ = 40 + 12 + 3(16) = 100 g/mol.
Let us convert mol to grams.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ }[/tex] [tex]\rightarrow \boxed{ \ mass = mol \times Mr \ }[/tex]
[tex]\boxed{ \ CaCO_3 \ }[/tex] [tex]\rightarrow \boxed{ \ mass = 0.08 \ moles \times 100 \ \frac{g}{mol} \ }[/tex]
Thus, the theoretical yield of CaCO₃ in grams equal to 8 grams.
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