Question

Lithium oxide is used to remove water from air according to the reaction:
${Li}_{2}O+H_{2}O\to 2LiOH$
If $72kg$ of water is to be removed and $35kg$ of ${Li}_{2}O$ is available. Which is a limiting reactant?

Answer 1

We have,

mass of water = 72 kg = 72000g

moles of water = $\frac{72000}{18}=4000moles$

mass of $L{i}_{2}O$ = 35 kg = 35000g

moles of $L{i}_{2}O$ = $\frac{35000}{30}=1166.67moles$

since, from the above equation, we have

moles of $L{i}_{2}O$ = moles of $H_{2}O$

since, moles of $L{i}_{2}O$ is less as compared to moles of water.

Hence, $L{i}_{2}O$ is limiting reagent.