Question

To process certain metal ores, reactions with carbon can sometimes generate the pure metal at high temperatures.

Consider the reaction of zinc oxide with carbon:

ZnO + C → Zn + CO ; ΔH = 240 kJ/mol of rxn

How much heat needs to be absorbed by the system (in kJ) to react 3121 g of ZnO with 351 g of C until the reaction is completed?

Answer 1

Answer on Question #81831, Chemistry / Organic Chemistry

To process certain metal ores, reactions with carbon can sometimes generate the pure metal at high temperatures.

Consider the reaction of zinc oxide with carbon:

$$\mathrm{ZnO} + \mathrm{C} \rightarrow \mathrm{Zn} + \mathrm{CO}; \ \Delta H = 240\ \mathrm{kJ/mol}\ \text{of rxn}$$

How much heat needs to be absorbed by the system (in kJ) to react 3121 g of ZnO with 351 g of C until the reaction is completed?

Solution

Find limiting reactant:

$$\mathrm{ZnO} + \mathrm{C} \rightarrow \mathrm{Zn} + \mathrm{CO}; \ \Delta H = 240\ \mathrm{kJ/mol}\ \text{of rxn}$$ $$n = m/M$$ $$n(\mathrm{ZnO}) = \frac{3121\ \mathrm{g}}{81\ \frac{\mathrm{g}}{\mathrm{mol}}} = 38.53\ \mathrm{mol}$$ $$n(\mathrm{C}) = \frac{351\ \mathrm{g}}{12\ \frac{\mathrm{g}}{\mathrm{mol}}} = 29.25\ \mathrm{mol}$$

According to equation mole ratio $n(\mathrm{ZnO}) : n(\mathrm{C}) = 1:1$, then $n(\mathrm{ZnO})$ should be equal to $n(\mathrm{C})$, but we can see that $n(\mathrm{ZnO}) > n(\mathrm{C})$, $38.53\ \mathrm{mol} > 29.25\ \mathrm{mol}$. So, C is a limiting reactant.

Find heat that needs to be absorbed by the system (in kJ) to react 3121 g of ZnO with 351 g of C until the reaction is completed by solving the proportion

$$\mathrm{ZnO} + \mathrm{C} \rightarrow \mathrm{Zn} + \mathrm{CO}; \ \Delta H = 240\ \mathrm{kJ/mol}\ \text{of rxn}$$ $$\frac{29.25\ \mathrm{mol}}{1\ \mathrm{mole}} = \frac{x\ \mathrm{kJ}}{240\ \mathrm{kJ}}$$ $$x = 7020\ \mathrm{kJ}$$

Answer: 7020 kJ

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