Question

Please answer these. It's part of Quantitave Chemistry. Calculating limiting raegents

Answer 1

The limiting reactant refers to the reactant that is in small amount. The extent of the reaction depends on the amount of the limiting reactant.

What is limiting reactant?

The term limiting reactant refers to the reactant that is in small amount, We shall now solve the questions individually.

1) The reaction equation is;

CaO + H2O → Ca(OH)2

Number of moles of CaO = 10 g/56 g/mol = 0.18 moles

Number of moles of H2O = 10 g/18 g/mol = 0.56

Since the reaction is 1:1, CaO is the limiting reactant

Mass of calcium hydroxide = 0.18 moles * 74 g/mol = 13.32 g

2) The reaction equation is;

Mg + Br2 → MgBr2

Number of moles of Mg = 1 g/24 g/mol = 0.042 moles

Number of moles of Br2= 5 g/160 g/mol = 0.031 moles

Since the reaction is 1:1, Br2 is the limiting reactant

Mass of MgBr2 =  0.031 moles * 184 g/mol = 5.7 g

3) The reaction equation is;

CuO + H2 → Cu + H2O

Number of moles of CuO= 2.00 g/80 g/mol = 0.025 moles

Number of moles of H2 = 1.00g/2 g/mol = 0.5 moles

Since the reaction is 1:1, CuO is the limiting reactant

Mass of Cu = 0.025 moles * 63.5 g/mol = 1.59 g

4) The reaction equation is;

2Na + F2 → 2NaF

Number of moles of Na = 2.30 g/23 g/mol = 0.10 moles

Number of moles of F2 =  2.85 g/38 g/mol = 0.075 moles

If 2 moles of Na react with 1 mole of F2

x moles of  Na reacts with 0.075 moles of F2

x = 0.15

Hence Na is the limiting reactant

Mass of NaF = 0.10 moles * 42 g/mol = 4.2 g

5) The equation of the reaction is;

Fe2O3 + 2Al → 2Fe + Al2O3

Number of moles of Fe2O3  =  8.00 g/160 g/mol = 0.05 moles

Number of moles of Al = 2.16 g/27 g/mol = 0.08 moles

If 1 mole of Fe2O3  reacts with 2 mole of Al

x moles of Fe2O3 reacts with 0.08 moles of Al

x = 0.04 moles

Hence, Fe2O3 is the limiting reactant

1 mole of Fe2O3  yileds 2 moles of Fe

0.05 moles yileds 0.05 moles  * 2 moles/1 mole = 0.1 moles

Mass of Fe = 0.1 moles * 56 g/mol = 5.6 g

6) The equation of the reaction is;

2Al + 3Cl2 → 2AlCl3

Number of moles of Al = 13.5 g/27 g/mol = 0.5 moles

Num ber of moles of Cl2 =  42.6 g /71 g/mol = 0.6 moles

If 2 moles of Al reacts with 3 moles of Cl2

x moles of Al reacts with 0.6 moles of Cl2

x = 0.4 moles

Al is the limiting reactant

Mass of AlCl3 = 0.4 moles * 133 g/mol = 53.2 g

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