Question

7. При восстановлении 5,1 г оксида трехвалентного металла водородом образовалось 2,7 г воды. Определите молярную массу эквивалента металла и молярную массу металла.

Answer 1

Me2O3+3H2=2Me+3H2O

1                              3

 

n(H2O)=2,7:18=0,15;n(Me2O3)=0,15:3=0,05; M(Me2O3)=5,1:0,05=102;

 

M(Me)=[102 -(16*3)]:2=27г/моль-алюминий;  Мэ(Al)=27^3=9г/моль

Answer 2

I will answer in detail. 7. When 5.1 g of oxide of a trivalent metal is reduced by hydrogen, 2.7 g of water is formed. Determine the molar mass of the metal equivalent and the molar mass of the metal.

To solve this problem, we need to write the chemical equation of the reaction and balance it. Let M be the symbol of the metal, then the equation is:

M₂O₃ + 3H₂ → 2M + 3H₂O

The equation is already balanced, so we can use the law of conservation of mass to find the mass of the metal. Since 2.7 g of water is formed, the mass of hydrogen that reacted is 2.7 x 0.111 = 0.3 g (using the percentage composition of water). Therefore, the mass of the oxide that reacted is 5.1 - 0.3 = 4.8 g. The mass of the metal in the oxide is 4.8 - 2.7 = 2.1 g. The mass of the metal equivalent is the mass of the metal that combines with one gram of oxygen, so we can find it by dividing the mass of the metal by the mass of oxygen in the oxide:

M_e = 2.1 / 1.6 = 1.3125 g

The molar mass of the metal equivalent is the mass of the metal equivalent multiplied by the valency of the metal, which is 3 in this case:

M_me = 1.3125 x 3 = 3.9375 g/mol

The molar mass of the metal is the same as the molar mass of the metal equivalent, since the metal is monatomic:

M_m = M_me = 3.9375 g/mol

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