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 

Order of filling levels and sublevels of electrons: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²3f¹⁴5d¹⁰6p⁶for an element with 5 electrons in the 3d energy level: full configuration: 1s²2s²2p⁶3s²3p³.