Question

Consider the reaction

KOH(s) + CO2(g) → KHCO3(s)


How many grams of KOH are required to react completely with 4.19 mol of CO2?

How many grams of KHCO3 are produced from the complete reaction of 83.1 g of KOH?

Answer 1

Answer on Question #43921 - Chemistry - Inorganic Chemistry

Question:

Consider the reaction:

$$\mathrm{KOH}(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{KHCO}_3(s)$$

1) How many grams of KOH are required to react completely with 4.19 mol of CO₂?

2) How many grams of KHCO₃ are produced from the complete reaction of 83.1 g of KOH?

Answer:

1) Based on the equation shows that the number of moles of carbon dioxide $\mathrm{CO}_{2}$ and potassium hydroxide KOH equally – 4.19 mol. Molar mass of potassium hydroxide tiles equals 56 g/mol.

$$\mathrm{m}(\mathrm{KOH}) = \mathrm{M}(\mathrm{KOH}) \cdot \mathrm{n}(\mathrm{KOH})$$ $$\mathrm{m}(\mathrm{KOH}) = 56 \cdot 4.19 = 234.64 \text{ (g)}$$

2) For solving the problem number 2 is necessary to make the proportion:

$$\begin{array}{ll} 83.1 \text{ g} & \times \text{g} \\ \mathrm{KOH}(s) + \mathrm{CO}_2(g) \rightarrow \mathrm{KHCO}_3(s) \\ 56 \text{ g/mol} & 100 \text{ g/mol} \end{array}$$ $$X = \frac{83.1 \cdot 100}{56} = 148.39 \text{ (g)}$$

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