Limiting reagent: The reagent that limits or determines the amount of product th

Question

Limiting reagent: The reagent that limits or determines the amount of product that can be formed in a reaction. Excessive reagent: The reagent that is present more than enough to react with a limiting reagent. Percent yield = (actual yield/theoretical yield) times 100 For each of these balanced equations, do the following. Identify the limiting reagent. Calculate the amount of the product formed in moles and grams. Calculate the number of moles of excess reagent remaining after the reaction. 2 H_2 + O_2 rightarrow 2 H_2O (5.0 mol, 2.3 mol) 4 P + 5 O_2 rightarrow 2 P_2O_5 (7 mol, 8 mol) 2A1 + 3 Cl_2 rightarrow 2 AlCl_3 (5.4 mol, 8.0 mol)

Explanation / Answer

Reaction 1

a) Limiting reactant

Write mole ratio between reactants

2 mol H2 : 1 mol O2

Calculate moles of O2 required to react with 5.0 mol H2 ( that is given in the problem)

n O2 = 5.0 mol H2 x 1 mol O2/ 2 mol H2

=2.5 mol O2

Compare this calculated mol of O2 with given.

Given moles of O2 are less than 2.5 so O2 is limiting reactant.

Calculation of amount of product formed:

Amount of product formed depends on the amount of limiting reactant present.

Product is H2O

Determine the mol ratio between product and limiting reactant.

1 mol O2 : 2 mol H2O

n H2O formed = 2.3 mol O2 x 2 mol H2O / 1 mol O2 = 4.6 mol H2O

mass of H2O formed = n H2O x molar mass of H2O

= 4.6 mol H2O x 18.0148 g/mol

=82.87 g H2O

Moles of excess reactant = Given amount of excess reactant – reacted amount of excess reactant.

Calculate reacted amount using mole ratio

Moles of H2 required to react with 2.3 mol O2

= 2.3 mol O2 x 2 mol H2/1 mol O2

=4.6 mol

Therefore excess moles of reactant = 5.0 – 4.6 = 0.4 mol H2

We can follow same process.

Reaction 2

Limiting reactant

Moles of O2 required to react with 7 mol P

= 7.0 mol P x 5 mol O2 / 4 mol P

=8.75 mol O2

8.75 mol O2 is greater than 8.0 mol so O2 is limiting reactant and P is excess reactant.

Mass of P2O5 formed

n P2O5 = 8.0 mol O2 x 2 mol P2O5 / 5 mol O2

=3.2 mol P2O5

Mass of P2O5 = 3.2 mol P2O5 x 141.943 g/mol

=454.2176 g P2O5

Excess moles of P

Moles of P required to react with 8.0 mol O2

= 8.0 mol O2 x 4 mol P / 5 mol O2

= 6.4 mol

Excess moles of P = 7.0 mol – 6.4 mol = 0.60 mol P

Reaction III

n Cl2 = 5.4 mol Al x 3 mol Cl2 / 2 mol Al2

=8.1 mol Cl2

8.1 mol Cl2 is higher than given moles of Cl2 so Cl2 is limiting reagent.

Moles of AlCl3 = 5.4 mol Al x 2 mol AlCl3/ 2 mol Al

=5.4 mol AlCl3

Mass of AlCl3 = 5.4 mol AlCl3 x 133.341 g/mol

= 720.0 g AlCl3

Excess moles of Al = 5.4 mol – ( 8.0 mol Cl2 x 2 mol Al / 3 mol Cl2 )

= 0.067 mol

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