Question

Question #19122

Answer 1

`"5.25 g"` of `"NaF"` will be produced (or `"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 `"8:2"`, or better said, a `"4:1"` mole ratio between `"HF"` and `"NaF"`.

This means that `"4 moles"` of `"HF"` will produce `"1 mole"` of `"NaF"`. Since you were given the number of `"HF"` moles, and told that `Na_2SiO_3` was not a limiting reagent, you can determine the number of `"NaF"` moles to be

`"0.5 moles HF" * ("1 mole NaF")/("4 moles HF") = "0.125 moles"`

`"NaF"` has a molar mass of `"42.0 g/mol"`, which means that the mass produced will be

`"0.125 moles" * ("42.0 g")/("1 mole") = "5.25 g NaF"`