Question

Hydrochloric acid and magnesium carbonate react according to the following equation:
2HCL+MgCO3--> Mgcl2+H2O+CO2
(A) when 0.84g of MgCO3 reacts with excess HCL,what is the mass of MgCL2 that will be formed?
(B) if the amount of MgCL2 obtained is only 0.750g,calculate the percentage yield.

Answer 1

Answer on Question #56617 - Chemistry - Inorganic Chemistry

Question:

Hydrochloric acid and magnesium carbonate react according to the following equation:

$$2 \mathrm{HCL} + \mathrm{MgCO_3} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2O} + \mathrm{CO_2}$$

(A) when 0.84g of MgCO₃ reacts with excess HCL, what is the mass of MgCl₂ that will be formed?

(B) if the amount of MgCl₂ obtained is only 0.750g, calculate the percentage yield.

Solution

$$2 \mathrm{HCl} + \mathrm{MgCO_3} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2O} + \mathrm{CO_2}$$

A) $\mathrm{M}(\mathrm{MgCO_3}) = 24 + 12 + 16 \times 3 = 84\ (\mathrm{g/mol})$

$$\begin{array}{l} n(\mathrm{MgCO_3}) = \mathrm{m/M} = 0.84 / 84 = 0.01\ (\mathrm{mol}) \\ n(\mathrm{MgCl_2}) = n(\mathrm{MgCO_3}) = 0.01\ \mathrm{mol} \\ \mathrm{M}(\mathrm{MgCl_2}) = 24 + 35.5 \times 2 = 95\ (\mathrm{g/mol}) \\ \mathrm{m}(\mathrm{MgCl_2}) = \mathrm{M \times n} = 95 \times 0.01 = 0.95\ (\mathrm{g}) \\ \end{array}$$

B) Yield = $\mathrm{m/m_t \times 100\% = 0.750 / 0.95 \times 100\% = 78.95\%}$

Answer

A) $\mathrm{m}(\mathrm{MgCl_2}) = 0.95\ \mathrm{g}$

B) Yield = 78.95%

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