Question

How many grams of NaF form when .508 mol of HF react with excess Na2SiO3?

Answer 1

2HF + Na2SiO3 ---->  2NaF +  H2SiO3   is the balanced equation.

2 moles of HF  form 2 moles of NaF

therefore 0.508 Mol  HF forms 0.508 mol of NaF

In grams this is 0.508( rel atomic mass of Na + RAM of F)
                        = 0.508 (22.99+19)
                        =  21.33 grams