# Question #19122

Jan 29, 2015

$\text{5.25 g}$ of $\text{NaF}$ will be produced (or $\text{5 g}$ if you round to 1 sig fig - the number of sig figs 0.5 has).

So, you have your balanced chemical equation, which I won't write again here. Notice that you have an $\text{8:2}$, or better said, a $\text{4:1}$ mole ratio between $\text{HF}$ and $\text{NaF}$.

This means that $\text{4 moles}$ of $\text{HF}$ will produce $\text{1 mole}$ of $\text{NaF}$. Since you were given the number of $\text{HF}$ moles, and told that $N {a}_{2} S i {O}_{3}$ was not a limiting reagent, you can determine the number of $\text{NaF}$ moles to be

$\text{0.5 moles HF" * ("1 mole NaF")/("4 moles HF") = "0.125 moles}$

$\text{NaF}$ has a molar mass of $\text{42.0 g/mol}$, which means that the mass produced will be

$\text{0.125 moles" * ("42.0 g")/("1 mole") = "5.25 g NaF}$