What mass of 95% pure \(CaCO_3\) will be required to neutralise 50 mL of 0.5 M \(HCl\) solution according to the following reaction?
\(CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) + 2H_2O(l)\)
[Calculate upto second place of decimal point]
\(CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) + 2H_2O(l)\)
[Calculate upto second place of decimal point]
- 1.25 g
- 1.32 g
- 3.65 g
- 9.50 g
The Correct Option is B
Solution and Explanation
To determine the mass of 95% pure \(CaCO_3\) required to neutralise 50 mL of 0.5 M \(HCl\), we need to follow these steps:
1. Understand the reaction: The chemical equation given is \(CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + 2H_2O\). From this, we see that 1 mole of \(CaCO_3\) reacts with 2 moles of \(HCl\).
2. Calculate moles of \(HCl\): Using the formula \(Moles = Molarity \times Volume\ (in\ L)\), we have:
\[0.5 \, \text{M} \times 0.050 \, \text{L} = 0.025 \, \text{moles of } HCl\]
3. Determine moles of \(CaCO_3\) required: From the stoichiometry of the reaction, \(2 \, \text{moles of} \, HCl\) require \(1 \, \text{mole of} \, CaCO_3\). Therefore:
\[0.025 \, \text{moles } HCl \times \frac{1 \, \text{mole } CaCO_3}{2 \, \text{moles } HCl} = 0.0125 \, \text{moles of } CaCO_3\]
4. Calculate mass of pure \(CaCO_3\): Using the molar mass of \(CaCO_3\) (100 g/mol), the mass is:
\[0.0125 \, \text{moles } \times 100 \, \text{g/mol} = 1.25 \, \text{g}\]
5. Adjust for purity: Since the sample is 95% pure, the actual mass required is:
\[1.25 \, \text{g} \div 0.95 = 1.32 \, \text{g}\]
The mass of 95% pure \(CaCO_3\) required is 1.32 g.
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Concepts Used:
Stoichiometry
Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products, leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of the products can be empirically determined, then the amount of the other reactants can also be calculated.
Stoichiometry helps us determine how much substance is needed or is present. Things that can be measured are;
- Reactants and Products mass
- Molecular weight
- Chemical equations
- Formulas
Stoichiometric Coefficient
The Stoichiometric coefficient of any given component is the number of molecules and/or formula units that participate in the reaction as written.
Mole Ratios
The mass of one mole of a substance in grams is called molar mass. The molar mass of one mole of a substance is numerically equal to the atomic/molecular formula mass.