Asked by omar
explain how the stoichiometry is predicted by the conservation of mass.
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Short answer: Stoichiometry follows from the law of conservation of mass because a balanced chemical equation simply counts atoms. Atoms are neither created nor destroyed in a reaction, so the coefficients in a balanced equation give the fixed mole ratios among reactants and products. Those mole ratios let you convert between amounts (moles or mass) of any species in the reaction.
How you use that in practice
1. Write and balance the chemical equation so every element has the same atom count on both sides. The coefficients are mole ratios.
2. Convert given masses to moles (moles = mass / molar mass).
3. Use the mole ratios from the balanced equation to find how many moles of product (or other reactant) are produced/consumed.
4. Convert moles back to mass if required.
5. If both reactants are present, identify the limiting reagent (the one that runs out first) by comparing the available mole ratio to the stoichiometric ratio.
6. Optional check: total mass of all products plus any unreacted material equals the total mass of the starting reactants (within rounding).
Worked example (CH4 combustion)
Balanced equation: CH4 + 2 O2 → CO2 + 2 H2O
Molar masses: CH4 = 16.04 g/mol, O2 = 32.00 g/mol, CO2 = 44.01 g/mol, H2O = 18.02 g/mol
Given 16.04 g CH4 and 48.00 g O2:
- Moles CH4 = 16.04 / 16.04 = 1.00 mol
- Moles O2 = 48.00 / 32.00 = 1.50 mol
Stoichiometry requires 2 mol O2 per 1 mol CH4, so O2 is the limiting reagent (1.50 mol O2 can react with 0.75 mol CH4).
- Moles CO2 produced = moles CH4 consumed = 0.75 mol → mass CO2 = 0.75 × 44.01 = 33.01 g
- Moles H2O produced = 2 × 0.75 = 1.50 mol → mass H2O = 1.50 × 18.02 = 27.03 g
- Unreacted CH4 = 1.00 − 0.75 = 0.25 mol → mass leftover CH4 = 0.25 × 16.04 = 4.01 g
Mass check: 33.01 + 27.03 + 4.01 ≈ 64.05 g, which equals the initial total mass 16.04 + 48.00 ≈ 64.04 g (difference due to rounding) — demonstrating conservation of mass.
Summary
- Conservation of mass → atoms conserved → balanced equation with mole ratios.
- Stoichiometry is simply using those mole ratios with molar masses to predict quantities of reactants or products.
How you use that in practice
1. Write and balance the chemical equation so every element has the same atom count on both sides. The coefficients are mole ratios.
2. Convert given masses to moles (moles = mass / molar mass).
3. Use the mole ratios from the balanced equation to find how many moles of product (or other reactant) are produced/consumed.
4. Convert moles back to mass if required.
5. If both reactants are present, identify the limiting reagent (the one that runs out first) by comparing the available mole ratio to the stoichiometric ratio.
6. Optional check: total mass of all products plus any unreacted material equals the total mass of the starting reactants (within rounding).
Worked example (CH4 combustion)
Balanced equation: CH4 + 2 O2 → CO2 + 2 H2O
Molar masses: CH4 = 16.04 g/mol, O2 = 32.00 g/mol, CO2 = 44.01 g/mol, H2O = 18.02 g/mol
Given 16.04 g CH4 and 48.00 g O2:
- Moles CH4 = 16.04 / 16.04 = 1.00 mol
- Moles O2 = 48.00 / 32.00 = 1.50 mol
Stoichiometry requires 2 mol O2 per 1 mol CH4, so O2 is the limiting reagent (1.50 mol O2 can react with 0.75 mol CH4).
- Moles CO2 produced = moles CH4 consumed = 0.75 mol → mass CO2 = 0.75 × 44.01 = 33.01 g
- Moles H2O produced = 2 × 0.75 = 1.50 mol → mass H2O = 1.50 × 18.02 = 27.03 g
- Unreacted CH4 = 1.00 − 0.75 = 0.25 mol → mass leftover CH4 = 0.25 × 16.04 = 4.01 g
Mass check: 33.01 + 27.03 + 4.01 ≈ 64.05 g, which equals the initial total mass 16.04 + 48.00 ≈ 64.04 g (difference due to rounding) — demonstrating conservation of mass.
Summary
- Conservation of mass → atoms conserved → balanced equation with mole ratios.
- Stoichiometry is simply using those mole ratios with molar masses to predict quantities of reactants or products.
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