Question

part b
for the reaction, calculate how many grams of the product form when 2.6 g of cao completely reacts.
assume that there is more than enough of the other reactant.
cao (s) + co₂ (g) → caco₃ (s)
express your answer using two significant figures.
part c
for the reaction, calculate how many grams of the product form when 2.6 g of mg completely reacts.
assume that there is more than enough of the other reactant.
2 mg (s) + o₂ (g) → 2 mgo (s)
express your answer using two significant figures.

Explanation:

Step1: Find moles of CaO

Molar mass of \( \text{CaO} \) is \( 40.08 + 16.00 = 56.08 \, \text{g/mol} \).
Moles of \( \text{CaO} = \frac{2.6 \, \text{g}}{56.08 \, \text{g/mol}} \approx 0.04636 \, \text{mol} \).

Step2: Use stoichiometry (1:1 ratio)

From reaction \( \text{CaO} + \text{CO}_2
ightarrow \text{CaCO}_3 \), moles of \( \text{CaCO}_3 \) = moles of \( \text{CaO} \approx 0.04636 \, \text{mol} \).

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

Molar mass of \( \text{CaCO}_3 \) is \( 40.08 + 12.01 + 3(16.00) = 100.09 \, \text{g/mol} \).
Mass = \( 0.04636 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 4.6 \, \text{g} \) (2 sig figs).

Answer:

\( \boxed{4.6} \) (grams, for Part B)

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Part C: Reaction \( 2\text{Mg} + \text{O}_2

ightarrow 2\text{MgO} \)

Step1: Moles of Mg

Molar mass of \( \text{Mg} = 24.31 \, \text{g/mol} \).
Moles of \( \text{Mg} = \frac{2.6 \, \text{g}}{24.31 \, \text{g/mol}} \approx 0.1069 \, \text{mol} \).

Step2: Stoichiometry (2:2 ratio, so 1:1)

Moles of \( \text{MgO} \) = moles of \( \text{Mg} \approx 0.1069 \, \text{mol} \).

Step3: Mass of \( \text{MgO} \)

Molar mass of \( \text{MgO} = 24.31 + 16.00 = 40.31 \, \text{g/mol} \).
Mass = \( 0.1069 \, \text{mol} \times 40.31 \, \text{g/mol} \approx 4.3 \, \text{g} \) (2 sig figs).