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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} K=(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=
`(2)/(2)* 0.57=0.57moles` of
`KOH`

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

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