A student wanted to make 15.0 g of zinc chloride. The equation for the reaction is:
[tex]\[ \text{ZnCO}_3 + 2 \text{HCl} \rightarrow \text{ZnCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \][/tex]

What mass of zinc carbonate should the student add to the hydrochloric acid to make 15.0 g of zinc chloride?

[tex]\[ \left[ A_r \text{ of C} = 12; \text{Zn} = 65; \text{O} = 16; \text{Cl} = 35.5; \text{H} = 1 \right] \][/tex]

A. 11.0 g
B. 13.8 g
C. 15.0 g
D. 22.0 g

[1 mark]



Answer :

To determine the mass of zinc carbonate (ZnCO₃) needed for the given reaction to produce 15.0 g of zinc chloride (ZnCl₂), follow these detailed steps:

1. Calculate the molar mass of ZnCl₂:
- Atomic masses (Ar): [tex]\( Zn = 65 \)[/tex], [tex]\( Cl = 35.5 \)[/tex]
- Molar mass of [tex]\(ZnCl_2\)[/tex]:
[tex]\[ 65 + 2 \times 35.5 = 65 + 71 = 136 \text{ g/mol} \][/tex]

2. Calculate the molar mass of ZnCO₃:
- Atomic masses (Ar): [tex]\( Zn = 65 \)[/tex], [tex]\( C = 12 \)[/tex], [tex]\( O = 16 \)[/tex]
- Molar mass of [tex]\(ZnCO_3\)[/tex]:
[tex]\[ 65 + 12 + 3 \times 16 = 65 + 12 + 48 = 125 \text{ g/mol} \][/tex]

3. Calculate the moles of ZnCl₂ needed:
- Desired mass of ZnCl₂: [tex]\( 15.0 \text{ g} \)[/tex]
- Moles of ZnCl₂:
[tex]\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{15.0 \text{ g}}{136 \text{ g/mol}} \approx 0.1103 \text{ mol} \][/tex]

4. Calculate the mass of ZnCO₃ needed:
- Moles of ZnCO₃ needed is equal to moles of ZnCl₂ (1:1 ratio in the reaction):
[tex]\[ \text{mass} = \text{moles} \times \text{molar mass of ZnCO}_3 = 0.1103 \text{ mol} \times 125 \text{ g/mol} \approx 13.79 \text{ g} \][/tex]

5. Determine the closest choice:
- Provided choices:
[tex]\[ A. \quad 11.0 \text{ g} \\ B. \quad 13.8 \text{ g} \\ C. \quad 15.0 \text{ g} \\ D. \quad 22.0 \text{ g} \][/tex]
- The calculated mass of [tex]\( 13.79 \text{ g} \)[/tex] is closest to choice B: [tex]\( 13.8 \text{ g} \)[/tex].

Answer: B [tex]\(13.8 \text{ g}\)[/tex]