Question

part a
for the reaction, calculate how many grams of the product form when 2.2 g of cao completely reacts.
assume that there is more than enough of the other reactant.
express your answer using two significant figures.
cao (s) + co₂ (g) → caco₃ (s)
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Explanation:

Step1: Calculate moles of CaO

Molar mass of \( \text{CaO} \) is \( 40.08 + 16.00 = 56.08 \, \text{g/mol} \). Moles of \( \text{CaO} = \frac{2.2 \, \text{g}}{56.08 \, \text{g/mol}} \approx 0.0392 \, \text{mol} \).

Step2: Relate moles of CaO to \( \text{CaCO}_3 \)

From the reaction \( \text{CaO}(s) + \text{CO}_2(g)
ightarrow \text{CaCO}_3(s) \), the mole ratio of \( \text{CaO} \) to \( \text{CaCO}_3 \) is \( 1:1 \). So moles of \( \text{CaCO}_3 = 0.0392 \, \text{mol} \).

Step3: Calculate mass of \( \text{CaCO}_3 \)

Molar mass of \( \text{CaCO}_3 \) is \( 40.08 + 12.01 + 3 \times 16.00 = 100.09 \, \text{g/mol} \). Mass of \( \text{CaCO}_3 = 0.0392 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 3.9 \, \text{g} \) (rounded to two significant figures).

Answer:

\( 3.9 \)