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

Sagot :

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

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

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

[tex]2K+2H_2O\rightarrow 2KOH+H_2[/tex]  

According to stoichiometry :

2 moles of [tex]K[/tex] produce = 2 moles of [tex]KOH[/tex]

Thus 0.57 moles of [tex]K[/tex] will produce=[tex]\frac{2}{2}\times 0.57=0.57moles[/tex]  of [tex]KOH[/tex]

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

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

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