Question

part a
for the reaction, calculate how many grams of the product form when 2.8 g of na₂o completely reacts.
assume that there is more than enough of the other reactant.
express your answer using two significant figures.
na₂o(s) + h₂o(l) → 2 naoh(aq)

Explanation:

Step1: Calculate moles of \( \ce{Na2O} \)

Molar mass of \( \ce{Na2O} \) is \( 2\times23 + 16 = 62 \, \text{g/mol} \). Moles of \( \ce{Na2O} = \frac{2.8 \, \text{g}}{62 \, \text{g/mol}} \approx 0.0452 \, \text{mol} \).

Step2: Relate moles of \( \ce{Na2O} \) to \( \ce{NaOH} \)

From the reaction \( \ce{Na2O + H2O -> 2NaOH} \), 1 mol \( \ce{Na2O} \) produces 2 mol \( \ce{NaOH} \). So moles of \( \ce{NaOH} = 2\times0.0452 \, \text{mol} = 0.0904 \, \text{mol} \).

Step3: Calculate mass of \( \ce{NaOH} \)

Molar mass of \( \ce{NaOH} \) is \( 23 + 16 + 1 = 40 \, \text{g/mol} \). Mass of \( \ce{NaOH} = 0.0904 \, \text{mol} \times 40 \, \text{g/mol} \approx 3.6 \, \text{g} \) (two significant figures).

Answer:

\( 3.6 \)