Question

Calculate the mass of water vapor that is produced by the reaction: 

Answer 1

Answer:

$m_{H_2O}=0.353gH_2O$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to identify the required limiting reactant by calculating the moles of water vapor produced by each reactant, CO2 and KOH, as shown below:

$1.4gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molH_2O}{1molCO_2}=0.0318mol H_2O\\\\2.2gKOH*\frac{1molKOH}{56.11gKOH}*\frac{1molH_2O}{2molKOH}=0.0196mol H_2O$

In such a way, since 2.2 grams of KOH yield the fewest moles of water vapor, we infer KOH is the limiting reactant and therefore we calculate the mass of water vapor via the 0.0196 moles we obtained:

$m_{H_2O}=0.0196molH_2O*\frac{18.02gH_2O}{1molH_2O}=0.353gH_2O$

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