Question

Limiting Reactions Question #11

Part a.

For the following reaction, 0.289 moles of hydrochloric acid are mixed with 0.562 moles of oxygen gas.

Hydrochloric acid (aq) + oxygen (g) = water (l) + chlorine (g)

What is the FORMULA for the limiting reagent   ________________ and what is the maximum amount of water that can be produces? _____________moles

Part b.

For the following reaction, 9.80 grams of butane (C4H10) are allowed to react with 19.0 grams of oxygen gas.

Butane (C4H10)(g) + oxygen (g) = carbon dioxide (g) + water

What is the maximum amount of carbon dioxide that can be formed? ______________grams

What is the FORMULA for the limiting reagent ______________ and what amount of excess reagent remains after the reaction is complete? _______________grams

Answer 1

Part a

Write the balanced chemical equation

4HCl (aq) + O2(g) ---- > 2Cl2 (g) + 2H2O (l)

Mol ratio between HCl : O2 is 4 : 1

Lets calculate mole of oxygen required to react with given moles of HCl

Mol of O2 = 0.289 mol HCl x 1 mol O2 / 4 mol HCl

0.289*1/4=0.0723 mol O2

Actual moles of O2 present = 0.562 so O2 are present in excess.

HCl is limiting reactant

Amount of water produced depends on the amount of limiting reactant

Mol ratio HCl to water is 4 : 2

Moles of water produced = moles of HCl x 2 mol H2O / 4 mol HCl

0.289*2/4=0.1445 moles of H2O

moles of water formed = 0.1445 mol

Part b

Write balanced reaction

2 C4H10+ 13O2 ---> 10 H2O + 8 CO2

Calculate moles of butane

Mol = mass / molar mass

Molar mass of butane is 58.12 g/mol

9.80/58.12=0.1686 mol butane

Calculate moles of O2

Molar mass of O2 is 31.998 g/mol

19.0/31.998=0.5938 mol O2

Lets calculate moles of CO2 using both reactants

Mol ratio between butane to CO2 is 2 : 8

O2 to CO2 is 13 : 8

Moles of CO2 using butane

=moles of butane x 8 mol CO2/2 mol butane

0.1686*8/2=0.6744 mol CO2

Moles of CO2 using O2

0.5938*8/13=0.3654 mol CO2

Moles of O2 gives less moles of CO2 so this is maximum yield.

Since O2 gives less moles of CO2 so O2 is the limiting reactant and butane is in excess

We can find mass of CO2

Molar mass of CO2 is 44.01 g/mol

Therefore maximum mass formed = moles x molar mass

0.3654*44.01=16.0813 g CO2

Calculation of reagent remained after reaction

Butane is the limiting reactant and that will remain after reaction.

Calculate moles of butane required to react with O2

Moles of butane = moles of O2 x 2 butane / 13 O2

0.5938*2/13=0.0914 moles of butane

Excess moles of butane = moles of butane present - moles of butane required in reaction

0.1686-0.0914=0.0772 moles of butane

Molar mass of butane is 58.12 g /mol

Mass of butane

0.0772*58.12=4.50 g

So the excess mass of butane is 4.50 g