How many grams of the excess reactant are left after the reaction is complete?

Question

The reaction of Cr₂O₃ with silicon metal at high temperatures will make chromium metal.

2Cr2​O3​(s)+3Si(s)4Cr(l)+3SiO2​(s)

 The reaction is begun with 137.00 g of Si and 125.00 g of Cr₂O₃.

Answer 1

2Cr₂​O₃​(s) + 3Si(s) ==> Cr(l) + 3SiO₂​(s) ... balanced equation

First, find the limiting reactant:

137.00 g Si x 1 mol Si / 28.09 g = 4.877 moles Si

125.00 g Cr₂​O₃ x 1 mol Cr₂​O₃ / 151.99 g = 0.8224 moles Cr₂​O₃

Based on mole ratio of 2:3 for Cr₂​O₃ : Si, the Cr₂​O₃ will be limiting

Excess reactant therefore is Si.

How much Si is used up? 0.8224 mols Cr₂​O₃ x 3 mols Si / 2 mols Cr₂​O₃ = 1.233 mols Si used

How many moles Si are left? 4.877 mols Si - 1.233 mols Si = 3.643 mols Si left over

Converting this to grams Si left over, we have...

3.643 mols Si x 28.09 g / mol = 102.3 g Si left over after the reaction