Questions: Practice - Stoichiometry 1) In a chemical reaction requiring two atoms of phosphorus for five atoms of oxygen, how many grams of oxygen are required by 3.10 g of phosphorus? 2) What mass of sulfur must combine with aluminum to form 600 lbs .(272.4 kg) of aluminum sulfide?

The mass of oxygen required is \(\boxed{4.0032 \, \text{g}}\).

For the second question, we need the molar masses of aluminum (Al) and sulfur (S):

• Aluminum (Al): 26.98 g/mol
• Sulfur (S): 32.07 g/mol

The formula for aluminum sulfide is \(\text{Al}_2\text{S}_3\).

The reaction between aluminum and sulfur is: \[ 2 \text{Al} + 3 \text{S} \rightarrow \text{Al}_2\text{S}_3 \]

Convert the mass of aluminum sulfide to moles: \[ \text{Molar mass of } \text{Al}_2\text{S}_3 = 2(26.98) + 3(32.07) = 150.14 \, \text{g/mol} \] \[ \text{Moles of } \text{Al}_2\text{S}_3 = \frac{272.4 \, \text{kg} \times 1000 \, \text{g/kg}}{150.14 \, \text{g/mol}} = 1814.3 \, \text{mol} \]

From the balanced equation, 1 mole of \(\text{Al}_2\text{S}_3\) requires 3 moles of sulfur. Therefore, the moles of sulfur required are: \[ \text{Moles of S} = 1814.3 \, \text{mol} \times \frac{3 \, \text{mol S}}{1 \, \text{mol } \text{Al}_2\text{S}_3} = 5442.9 \, \text{mol S} \]

Now, calculate the mass of sulfur required: \[ \text{Mass of S} = 5442.9 \, \text{mol} \times 32.07 \, \text{g/mol} = 174,500 \, \text{g} = 174.5 \, \text{kg} \]

The mass of sulfur required is \(\boxed{174.5 \, \text{kg}}\).