Question

If there is sufficient water in the reaction system, how many grams of KOH can be produced from 22.2 g of K?

Answer 1

Answer: 31.9 g of KOH can be produced from 22.2 g of KOH

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} K=\frac{22.2g}{39g/mol}=0.57moles$  

$2K+2H_2O\rightarrow 2KOH+H_2$  

According to stoichiometry :

2 moles of $K$ produce = 2 moles of $KOH$

Thus 0.57 moles of $K$ will produce=$\frac{2}{2}\times 0.57=0.57moles$  of $KOH$

Mass of $KOH=moles\times {\text {Molar mass}}=0.57moles\times 56g/mol=31.9g$

Thus 31.9 g of KOH can be produced from 22.2 g of KOH