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)
provide feedback

Explanation:

Step1: Calculate moles of $\ce{Na2O}$

Molar mass of $\ce{Na2O}$: $2\times22.99 + 16.00 = 61.98\ \text{g/mol}$.
Moles of $\ce{Na2O} = \frac{2.8\ \text{g}}{61.98\ \text{g/mol}} \approx 0.0452\ \text{mol}$.

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

From reaction: $1\ \text{mol}\ \ce{Na2O}
ightarrow 2\ \text{mol}\ \ce{NaOH}$.
Moles of $\ce{NaOH} = 0.0452\ \text{mol} \times 2 = 0.0904\ \text{mol}$.

Step3: Calculate mass of $\ce{NaOH}$

Molar mass of $\ce{NaOH}$: $22.99 + 16.00 + 1.008 = 39.998\ \text{g/mol}$.
Mass of $\ce{NaOH} = 0.0904\ \text{mol} \times 39.998\ \text{g/mol} \approx 3.6\ \text{g}$ (2 sig figs).

Answer:

$\boxed{3.6}$