# How many grams of NaF form when 25.7g of HF reacts with excess sodium silicate?

Jan 12, 2016

$2 H F \left(a q\right) + N {a}_{2} S i {O}_{3} \left(a q\right) \rightarrow 2 N a F \downarrow + {H}_{2} S i {O}_{3} \left(a q\right)$
The stoichiometry of the equation is crucial; inasmuch as the stoichiometry, the chemical proportion, shows that hydrogen fluoride and sodium fluoride are formed in equal amounts, and all I need are the molecular weights of hydrogen fluoride (20.0*g*mol^-1) and sodium fluoride (42.0*g*mol^-1).
So moles of $H F$ $=$ $\frac{25.7 \cdot g}{20.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $m o l$
Moles of $N a F$ $=$ $\frac{20.0 \cdot g}{25.7 \cdot g \cdot m o {l}^{-} 1} \times 42.0 \cdot g \cdot m o {l}^{-} 1$ $=$ ?? $g$