Answer to Question #168379 in Chemistry for Emily

Question

In the following reaction 12.60g of CS2 burns with 5.80g of 02? Calculate the number of moles CS2 and O2 before the reaction occurs. Determine the limiting and the excess reactants, how many grams of CO2 would be produced. If 2.05g of CO2 was produced in lab, what is the percent yield. How many grams would be left over of the excess reactant

Answer 1

The reaction can be described as following:

CS₂ + 3 O₂ = CO₂ + 2 SO₂

The number of moles of CS₂:

n(CS₂) = m(CS₂) / Mr(CS₂) = 12.60 g / 76.14 g/mol = 0.1655 mol

n(O₂) = m(O₂) / [3 × Mr(O₂)] = 5.80 g / [3 × 32.0 g/mol] = 0.060 mol

From here, O₂ is the limiting reactant, CS₂ is the excess reactant.

The weight of CO2 produced:

m(CO₂) = n(O₂) × Mr(CO₂) = 0.060 mol × 44.01 g/mol = 2.64 g

The percent yield of carbon dioxide equals:

% w = [m(actual) / m(theoretical)] × 100 % = [2.05 g / 2.64 g] × 100 % = 77.65 %

Answer: 77.65 %