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bagirrra123 [75]
2 years ago
14

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

Chemistry
1 answer:
Arada [10]2 years ago
4 0

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.

<h3>What is limiting reactant?</h3>

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

Leran more about limiting reactants: brainly.com/question/14225536

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