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 125 g of potassium is added to 65 g of water?

• A. 1.61 g
• B. 3.21 g
• C. 4.73 g
• D. 6.44 g

Answer 1

The correct option is B 3.21 g

Moles of $K$ = $\frac{\text{given mass}}{\text{molar mass}}=\frac{125}{39}=3.21$ moles
Moles of $H_{2}O=\frac{\text{given mass}}{\text{molar mass}}=\frac{65}{18}=3.61$ moles
According to the reaction, 2 moles of $K$ require 2 moles of $H_{2}O.$
Therefore, 3.21 moles of $K$ will require 3.21 moles of $H_{2}O$
From the above calculations we can see that 0.40 moles of $H_{2}O$ is in excess.
Therefore, potassium is the limiting reagent.
Now, 2 moles of $K$ produce 1 mole of $H_2$
3.21 moles of $K$ will produce $\frac{1}{2}\times 3.21=1.605$ moles of $H_2$
So, mass of $H_2$ produced $=(1.605\times 2)$ g
= $3.21$ g