Question

Like most alkali metals, potassium reacts with water to form basic potassium hydroxide and hydrogen gas according to the reaction,
$2K+2H_{2}O\to 2KOH+H_2$
What amount of hydrogen gas will be formed when 117 g of potassium is added to 72 g of water?

Answer 1

Moles of $K$ = $\frac{\text{given mass}}{\text{molar mass}}=\frac{117}{39}=3moles$
$Molesof{H}_{2}O=\frac{\text{given mass}}{\text{molar mass}}=\frac{72}{18}=4$ moles
According to the reaction, 2 moles of $K$ require 2 moles of $H_{2}O.$
Therefore, 3 moles of $K$ will require 3 moles of $H_{2}O$
From the above calculations we can see that 1 mole of $H_{2}O$ is in excess.
Therefore, potassium is the limiting reagent.
Now, 2 moles of $K$ produce 1 mole of $H_2$
3 moles of $K$ will produce $\frac{1}{2}\times 3=1.5$ moles of $H_2$
So, mass of $H_2$ produced $=(1.5\times 2)$ g
= $3.0$ g