As a result of the interaction of 22 g of higher oxide of the element of the main subgroup of group IV with water, 31 g of acid was formed. Set the name of an unknown chemical element.

Carbon (C)

Explanation:

Because the element is the element of the main subgroup of group IV , it has oxidation number +4 in the oxide, and oxide formula is EO2.

reaction:

                                     EO2 + H2O ---> H2EO3

from reaction               1 mol    1 mol       1 mol

31 g acid - 22 g oxide = 9 g water

M(H2O) = 18 g/mol

9g * 1 mol/18 g = 0.5 mol H2O

As we see from reaction molar ratios EO2 : H2O : H2EO3 = 1 : 1 : 1,

so if  we have 0.5 mol H2O , we also have 0.5 mol EO2.

0.5 mol EO2 has mass 22 g.

Molar mass(EO2) = mass/ number of moles = 22 g/ 0.5 mol = 44 g/mol

Molar mass (EO2) = M(E) + 2M(O) = 44 g/mol

M(E) + 2 *16 g/mol = 44 g/mo

M(E) = 44-32 = 12 g/mol

Molar mass 12 g/ mol is molar mass of carbon.

So, element is carbon (C).

the answer for sodium is 0.186

the answer for rubidium is 0.251