Question

3. Too make exactly 100. grams of ferric sulfate how many grams of sulfuric acid would be

needed?

4. If you had 36.2 grams of aluminum metal reaction with 45.0 grams of copper (II) hydroxide,

which is the limiting and which is the excess reactant?

Answer 1

3. Write the equation:

$$2 \text{FeSO}_4 + \text{H}_2\text{SO}_4 + \text{H}_2\text{O}_2 → \text{Fe}_2\text{(SO}_4)_3 + 2 \text{H}_2\text{O}$$

We need 100 g of ferric sulfate, which corresponds to the following amount of substance:

$$n_1=\frac{m_1}{M_1}=\frac{100}{56·2+(32+16·4)·3}=0.25\text{ mol}.$$

From the reaction we notice that one molecule of the sulfate is produced by one molecule of the acid, that is, $n_1=n_2$, so, the mass of the sulfuric acid is

$$m_2=n_2M_2=0.25·(1·2+32+16·4)=24.5\text{ g}.$$

4. The reaction:

$$2\text{Al} +3\text{Cu(OH)}_2 → 3\text{Cu} + 2\text{Al(OH)}_3.$$

Find the amount of substance of each:

$$n_1=\frac{m_1}{M_1}=\frac{36.2}{27}=1.34\text{ mol}.\\\space\\ n_2=\frac{m_2}{M_2}=\frac{45}{64+(16+1)·2}=0.459\text{ mol}.\\\space\\$$

From the reaction we notice the number of aluminum atoms and the hydroxide molecules. Divide the corresponding amount of substance by the number of elemetns:

$$L_1=\frac{n_1}{N_1}=\frac{1.34}{2}=0.67\text{ mol/Al atom},\\ L_2=\frac{n_2}{N_2}=\frac{0.459}{3}=0.153\text{ mol/Cu(OH)}_2\text{ molecule},\\ L_2<L_1,$$

the hydroxide is the limiting one.