How many grams of iron (II) oxide can be produced from 3.4 g of iron in this balanced equation?
4Fe +3022Fe₂O3
- How many grams of magnesium carbonate is produced from 6.7 g of magnesium oxide in this
balanced equation?
MgO + CO₂ → MgCO3 - How many grams of calcium hydroxide is produced from 9.4 g of calcium chloride in this balanced
equation?
The Correct Answer and Explanation is :
To answer your questions, we need to balance the chemical equations, determine the molar masses of the reactants and products, and use stoichiometry to find the mass of the products formed from the given masses of the reactants.
1. Iron (II) oxide from iron
Balanced Equation:
[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 ]
Step 1: Calculate moles of Fe.
- Molar mass of Fe = 55.85 g/mol
- Moles of Fe from 3.4 g:
[ \text{Moles of Fe} = \frac{3.4 \text{ g}}{55.85 \text{ g/mol}} \approx 0.061 \text{ mol} ]
Step 2: Use stoichiometry to find moles of Fe₂O₃ produced.
From the balanced equation, 4 moles of Fe produce 2 moles of Fe₂O₃:
[ \text{Moles of Fe}_2\text{O}_3 = \frac{0.061 \text{ mol Fe}}{4} \times 2 \approx 0.0305 \text{ mol Fe}_2\text{O}_3 ]
Step 3: Calculate grams of Fe₂O₃.
- Molar mass of Fe₂O₃ = 159.69 g/mol
[ \text{Grams of Fe}_2\text{O}_3 = 0.0305 \text{ mol} \times 159.69 \text{ g/mol} \approx 4.87 \text{ g} ]
2. Magnesium carbonate from magnesium oxide
Balanced Equation:
[ \text{MgO} + \text{CO}_2 \rightarrow \text{MgCO}_3 ]
Step 1: Calculate moles of MgO.
- Molar mass of MgO = 40.30 g/mol
- Moles of MgO from 6.7 g:
[ \text{Moles of MgO} = \frac{6.7 \text{ g}}{40.30 \text{ g/mol}} \approx 0.166 \text{ mol} ]
Step 2: Use stoichiometry to find moles of MgCO₃ produced.
From the balanced equation, 1 mole of MgO produces 1 mole of MgCO₃:
[ \text{Moles of MgCO}_3 = 0.166 \text{ mol MgO} ]
Step 3: Calculate grams of MgCO₃.
- Molar mass of MgCO₃ = 84.31 g/mol
[ \text{Grams of MgCO}_3 = 0.166 \text{ mol} \times 84.31 \text{ g/mol} \approx 14.0 \text{ g} ]
3. Calcium hydroxide from calcium chloride
Balanced Equation:
[ \text{CaCl}_2 + 2 \text{NaOH} \rightarrow \text{Ca(OH)}_2 + 2 \text{NaCl} ]
Step 1: Calculate moles of CaCl₂.
- Molar mass of CaCl₂ = 110.98 g/mol
- Moles of CaCl₂ from 9.4 g:
[ \text{Moles of CaCl}_2 = \frac{9.4 \text{ g}}{110.98 \text{ g/mol}} \approx 0.0846 \text{ mol} ]
Step 2: Use stoichiometry to find moles of Ca(OH)₂ produced.
From the balanced equation, 1 mole of CaCl₂ produces 1 mole of Ca(OH)₂:
[ \text{Moles of Ca(OH)}_2 = 0.0846 \text{ mol CaCl}_2 ]
Step 3: Calculate grams of Ca(OH)₂.
- Molar mass of Ca(OH)₂ = 74.09 g/mol
[ \text{Grams of Ca(OH)}_2 = 0.0846 \text{ mol} \times 74.09 \text{ g/mol} \approx 6.26 \text{ g} ]
Summary of Results
- Iron (II) oxide produced from 3.4 g of iron: 4.87 g of Fe₂O₃
- Magnesium carbonate produced from 6.7 g of magnesium oxide: 14.0 g of MgCO₃
- Calcium hydroxide produced from 9.4 g of calcium chloride: 6.26 g of Ca(OH)₂
Explanation
The process of determining the mass of products formed from given reactants is based on stoichiometry, which involves balanced chemical equations that reflect the conservation of mass. The molar masses of the compounds are essential to convert grams to moles, allowing us to use the ratios from the balanced equations to predict the amounts of products produced. Each step ensures accurate conversion and results in the final calculations, showcasing the interrelationship between reactants and products in chemical reactions.