What mass of potassium hydroxide is formed when 8.2 g of potassium oxide is added to 1.3 g of water? answer key?

The reaction occurs by the following equation: 

K20 + H20 = 2KOH 

According to the equation we have following stoichiometric ratio:

n(K2O) : n(H20) = 1 : 1

From it, we can conclude that K2O and water react in same mass ratio.

Now, we have to determine limiting reagent (the one who determines how much the product will arise).

n(K2O)=m/M=8.2/94=0.09 mole

n(H20)=m/M=1.3/18=0.07 mole

So, there are fewer moles of H2O and it will determine the amount of the product.

n(H2O) : n(K0H) = 1 : 2

n(K0H)=n(H2O) x 2 = 0.17 mole

Finally, the was of KOH is:

m(KOH)= n x M = 0.17 x 56= 9.5 g 

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