For the following reaction, 5.43 grams of potassium hydroxide are mixed with exc

Question

For the following reaction, 5.43 grams of potassium hydroxide are mixed with excess carbon dioxide. The reaction yields 6.09 grams of potassium carbonate.


carbon dioxide (g) + potassium hydroxide (aq) potassium carbonate (aq) + water (l)
% yield=
theoretical=

Explanation / Answer

CO2(g) + 2 KOH(aq) ====> K2CO3 + H2O Moles = mass/molar mass Moles of KOH = 5.43 g / 56.11 g/mol =0.0964 mol KOH Now the molar ratio is 2 KOH : 1 K2CO3 0.0964 mol KOH * (1 mol K2CO3 / 2 mol KOH)*(138.21 g K2CO3 / 1mol K2CO3) 26.65 g K2CO3 [theoretical yield] % yield = (actual yield/theoretical yield)*100 = (6.09/26.65)*100 = 22.85 %

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