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Leni [432]
3 years ago
8

A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i

s the percent yield?
Chemistry
1 answer:
In-s [12.5K]3 years ago
3 0

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>

<em />

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

<em>Moles CaCl₂ = Moles CaCO₃:</em>

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

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