Chapter 9, Problem 53A

(a)

Interpretation Introduction

Interpretation:

The number of moles of the product produced by complete reaction of 1.00 g ammonia needs to be determined.

  $N{H}_3\left(g\right)+HCl\left(g\right)\to N{H}_{4}Cl\left(s\right)$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Explanation of Solution

The given reaction is as follows:

  $N{H}_3+HCl\to N{H}_{4}Cl$

The molar mass of Ammonia (17.031 g/mol)

If there's 1.00 g of Ammonia to react, it can be converted into number of moles.

1 mole of ammonia give 1 mol of product (as per equation). Thus, the number of moles of product formed will be:

  $\left(1.00gN{H}_3\right)\left(\frac{1mol{\text{ NH}}_3}{17.031g{\text{ NH}}_3}\right)\left(\frac{1mol{\text{ NH}}_{4}Cl}{1mol{\text{ NH}}_3}\right)=0.05872mol{\text{ NH}}_{4}Cl$

(b)

Interpretation Introduction

Interpretation:

The number of moles of the product produced by complete reaction of 1.00 g calcium oxide needs to be determined.

  $CaO\left(s\right)+C{O}_2\left(g\right)\to CaC{O}_3\left(s\right)$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Explanation of Solution

The given reaction is as follows:

  $CaO+C{O}_2\to CaC{O}_3$

The molar mass of CaO is 56.077 g/mol.

If there's 1.00 g of Calcium Oxide to react, the number of moles can be calculated from molar mass of CaO and the number of moles of product formed will be:

  $\left(1.00gCa\right)\left(\frac{1mol\text{ CaO}}{56.077g\text{ CaO}}\right)\left(\frac{1mol{\text{ CaCO}}_3}{1mol\text{ CaO}}\right)=0.018mol{\text{ CaCO}}_3$

(c)

Interpretation Introduction

Interpretation:

The number of moles of the product produced by complete reaction of 1.00 g sodium needs to be determined.

  $4Na\left(s\right)+O_2\left(g\right)\to 2N{a}_{2}O\left(s\right)$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Explanation of Solution

The molar mass of Sodium (22.99 g/mol)

If there's 1.00 g of Sodium to react it can be converted into number of moles.

From the reaction ratio of Na to Na₂O is 4:2 thus, number of moles of product formed will l be:

  $\left(1.00gNa\right)\left(\frac{1mol\text{ Na}}{22.99g\text{ Na}}\right)\left(\frac{2mol{\text{ Na}}_{2}O}{4mol\text{ Na}}\right)=0.022mol{\text{ NH}}_{4}Cl$

(d)

Interpretation Introduction

Interpretation:

The number of moles of the product produced by complete reaction of 1.00 g phosphorus needs to be determined.

  $2P\left(s\right)+3C{l}_2\left(g\right)\to 2PC{l}_3\left(l\right)$

Concept Introduction:

If complete conversion of any reactant takes place in the chemical reaction, it is considered as a limiting reactant.

Explanation of Solution

The given reaction is as follows:

  $2P\left(s\right)+3C{l}_2\left(g\right)\to 2PC{l}_3\left(l\right)$

The molar mass of Phosphorus (30.97 g/mol)

If there's 1.00 g of Phosphorus to react, it can be converted into number of moles.

The number of moles of product formed will be:

  $\left(1.00gP\right)\left(\frac{1mol\text{ P}}{30.97g\text{ P}}\right)\left(\frac{2mol{\text{ PCl}}_3}{2mol\text{ P}}\right)=0.03229mol{\text{ PCl}}_3$