Question

Lithium oxide, $\left(L{i}_{2}O\right)$ is used to aboard the space shuttle to remove water from the air supply according to the equation:
$L{i}_{2}O\left(s\right)+H_{2}O\left(g\right)⟶2LiOH\left(s\right)$
If there are $80.0$ kg of water to be removed and $65$ kg of $L{i}_{2}O$. How many kilograms of the excess reactant remain? $(Li=7,O=16,H=1)$

Answer 1

$30$ grams of lithium oxide require $18$ grams of water as per the above balanced equation.
$65$ kg of lithium oxide require $39$ kg of water.
Hence lithium oxide is limiting reagent.
Amount of water left would be $80-39=41$ kg.