What is the greatest amount of silicon tetrafluoride in grams that can be produced 48 grams of HF and 92 grams of SiO2? Which reactant is limiting? more in the description

Question

How many grams of the excess reactant will be left after the reaction?SiO2 + 4 HF  SiF4 + 2 H2O

Answer 1

(48g HF) * (1 mole HF / 20.0g HF) * (1 mole SiF₄ / 4 mole HF) = 0.60 mole SiF₄.

(92g SiO₂) * (1 mole SiO₂ / 60.1g SiO₂) * (1 mole SiF₄ / 1 mole SiO₂) = 1.53 mole SiF₄. HF is limiting.

(0.60 mole SiF₄) * (104.1g SiF₄ / 1 mole SiF₄) = 62g SiF₄.

Excess (0.93 mole SiF₄) * (1 mole SiO₂ / 1 mole SiF₄) * (60.1g SiO₂ / 1 mole SiO₂) = 56g SiO₂.