Question

Given the reaction:
$H_{2}S{O}_4\left(aq\right)+2LiOH\left(aq\right)\to L{i}_{2}S{O}_4\left(aq\right)+2H_{2}O\left(l\right)$
If you have 140 g of $H_{2}S{O}_4$ and 65 g of LiOH, calculate the mass of water produced?

• A. 48.78 g
• B. 26.62 g
• C. 55.39 g
• D. 38.15 g

Answer 1

The correct option is A 48.78 g

Moles of $H_{2}S{O}_4=\frac{140}{98}=1.43$ moles
Moles of LiOH = $\frac{65}{24}=2.71$ moles
$H_{2}S{O}_4\left(aq\right)+2LiOH\left(aq\right)\to L{i}_{2}S{O}_4\left(aq\right)+2H_{2}O\left(l\right)$
1 mole of $H_{2}S{O}_4$ reacts with 2 moles of LiOH
1.43 moles of $H_{2}S{O}_4$ will react with $1.43\times 2=2.86$ moles of LiOH
Hence, LiOH is the limiting reagent.
Now, 2 moles of LiOH produce 2 moles of water
So, 2.71 moles will produce 2.71 moles of water
Therefore, the mass of water produced $=2.71\times 18$ g
Mass of water produced = 48.78 g