Question

1)Use the balanced chemical equation below to calculate the mass of KCl that is formed for every 164 L of O2(g) produced. You may assume the reaction takes place at STP.

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

2)Identify the limiting reactant and calculate the theoretical yield of CO2(g), in grams, if 21.5 L of C3H8(g) reacts with 96.1 L of O2(g) according the the balanced chemical equation below.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

Answer 1

1) Balanced chemical equation :- 2 KClO3(s) → 2 KCl(s) + 3 O2(g) . In terms of moles we can say that for every 2 moles of KClO3 decomposition we get 2 moles of KCl and 3 moles of O2 or we can say that for every 3 moles of O2 produced we get 2 moles of KCl.

At STP(Standard Temperature and Pressure), volume of one mole of gas is 22.4 Litres.

So Number of moles of gas = Given Volume(in L) / 22.4

Hence Number of moles of O2 = 164/22.4 = 7.321 moles

For above equation,by unitary method, for every 1 mole of O2 produced we get 2/3 moles of KCl. So for 7.321 moles of O2 we get 7.321*(2/3) moles of KCl.

Moles of KCl formed = 7.321*(2/3) = 4.881 moles

Mass of KCl formed = Moles of KCl * Molar Mass of KCl = 4.881*74.55 = 363.88 g

Thereofore 363.88g of KCl formed.

2) At STP(Standard Temperature and Pressure), volume of one mole of gas is 22.4 Litres.

So Number of moles of gas = Given Volume(in L) / 22.4

Moles of C3H8 = 21.5/22.4 = 0.9598 moles, Moles of O2 = 96.1/22.4 = 4.290 moles

Therefore O2(g) is limiting reagent and theoretical yield of CO2(g) is 113.256g