Understanding Limiting Reactants
In chemical reactions, a limiting reactant is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant, as the reaction cannot continue without it.
The Reaction Equation
The reaction between magnesium (Mg) and oxygen ($O_2$) is: $$2Mg + O_2 \rightarrow 2MgO$$
Step-by-Step Problem Solving
1. Calculate Moles:
- Atomic mass of Mg = 24 g/mol. Moles of Mg = $\frac{2g}{24g/mol} = 0.0833$ mol.
- Molar mass of $O_2$ = 32 g/mol. Moles of $O_2$ = $\frac{3g}{32g/mol} = 0.09375$ mol.
2. Identify Limiting Reactant: From the equation, 2 moles of Mg require 1 mole of $O_2$. For 0.0833 mol of Mg, we need $\frac{0.0833}{2} = 0.04165$ mol of $O_2$. Since we have 0.09375 mol of $O_2$, which is greater than 0.04165 mol, Mg is the limiting reactant.
3. Moles Left Over:
- $O_2$ consumed = 0.04165 mol.
- $O_2$ left over = $0.09375 - 0.04165 = 0.0521$ mol.
4. Mass of MgO Produced:
- Moles of MgO produced = moles of Mg used = 0.0833 mol.
- Molar mass of MgO = $24 + 16 = 40$ g/mol.
- Mass = $0.0833 \times 40 = 3.33$ g.
5. Neutralization with $H_2SO_4$: Equation: $MgO + H_2SO_4 \rightarrow MgSO_4 + H_2O$
- Mole ratio is 1:1. So, 0.0833 mol of $H_2SO_4$ is required.
- Molar mass of $H_2SO_4 = 2(1) + 32 + 4(16) = 98$ g/mol.
- Mass required = $0.0833 \times 98 = 8.16$ g.