How many grams of the excess reagent are left over when 6.00 g of CS₂ gas react with 10.0 g of Cl₂ gas in the following reaction? CS₂(g)+ 3 Cl₂(g)→ CCl₄(l)+ S₂Cl₂(l) A)2.42 g B)2.77 g C)3.58 g D)4.00 g

Question

Quizplus · Accepted Answer

First, calculate the moles of each reactant: CS2 (6.00 g / 76.14 g/mol) = 0.0788 mol, Cl2 (10.0 g / 70.90 g/mol) = 0.141 mol. The reaction requires 3 moles of Cl2 per mole of CS2, so 0.0788 mol CS2 would need 0.2364 mol Cl2. Since only 0.141 mol Cl2 is available, Cl2 is the limiting reagent. Calculate the excess CS2: 0.0788 mol - (0.141 mol / 3) = 0.0311 mol CS2 left. Convert to grams: 0.0311 mol * 76.14 g/mol = 2.42 g.

How Many Grams of the Excess Reagent Are Left Over