( 1)2M+2H2O=2MOH+H2

(2)M2O+H2O=2MOH

After the reaction, the mass increment is 1.79- 1.40=0.39g, assuming that the mass of M2O is zero, and 0.39g is the increment of OH-, we can get n (moh) min = 0.39/17 = 0.023mol, assuming that the mass of M is zero. If n (MOH) max = (0.39/18) * 2 = 0.043mol, the molecular weight of MOH can be calculated from 1.79/n, so the molecular weight range of MOH is1.79/0.043 ~/kloc-.

In the question:

The mass mK of k is x, and the quantity nK of k is X/39.

The mass mK2O of K2O is (1.40-X), and the mass nK2O of K2O is (1.40-X)/(2*39+ 16).

List the equation for k:

nK+2nK2O = nKOH = 1.79/(39+ 16+ 1)

Substitute X= 1. 14.

So k is 1. 14g, and K2O is 0.26g

Very detailed ~ ~ ~