Practice #5- limiting reactant in grams

Question

Which is the greatest amount of silicon tetraflouride in grams that can be produced 48 grams of HF and 92 grams of SiO2? Which reactant is limiting? How many grams of the excess reactant will be left after the reaction?

SiO2 + 4HF = SiF4 + 2H2O

Answer 1

Hello, Vian,

Styart by calculating the number of moles of both reactants. Divide their masses by their molar masses:

SiO₂: (92 grams)/(60.1 grams/mole) = 1.53 moles SiO₂

HF: (48 grams)/(20.0 grams/mole) = 2.40 moles HF

The balanced equation says we need 4 moles of HF for every 1 mole SiO₂, a molar ratio of 4 to 1. Since we have 1.53 moles of SiO₂, four times that amount will exceed the amount of HF that would be needed to completely react. SiO₂ is therefore the limiting reagent. When all 2.40 moles of HF are reacted, they will have consumed (2.40 moles/4) = 0.60 moles of SiO₂. That will leave behind (1.53 - 0.931 = 0.931 moles of SiO₂. Multiply by the molar mass of SiO₂ to find grams left behind (0.931 moles)*(60.1 grams/mole) = 55.9 grams SiO₂ , left sitting in the beaker.

The equation says we'll obtain 1 mole of SiF₄ for each mole of SiO₂ consumed (or 1/4 of the moles consumed of HF). That gives us 0.600 moles of SiF₄. That means we'll have (0.60 moles)*(159.4 grams/mole) = 62.4 grams of SiF₄.

Bob