Question

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

Answer 1

`2HF(aq) + Na_2SiO_3(aq) rarr 2NaFdarr + H_2SiO_3(aq)`

Explanation:

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 `HF` `=` `(25.7*g)/(20.0*g*mol^-1)` `=` `??` `mol`

Moles of `NaF` `=` `(20.0*g)/(25.7*g*mol^-1) xx42.0*g*mol^-1` `=` `??` `g`