Question

При обработке серной кислотой оксида алюминия, масса солиоказалась равной 2,565 г. Какова масса оксида вступила в реакцию, если выход соли составил 75% от теоретически возможного? ОТВЕТ: 1,02 г.

Answer 1

Al2O3 + 3H2SO4 = Al2(SO4)3 + 3H2O
M(Al2O3) = 54+48 = 102 г/моль
M(Al2(SO4)3 = 342 г/моль
m(Al2(SO4)3 (теоретически) = 2,565 / 0,75 = 3,42 г
n(AL2(SO4)3 = 3,42 / 342 = 0,01 моль
n(Al2O3) = n(Al2(SO4)3 =0,01 моль
m(Al2O3) = 0,01*102 = 1,02 г

Answer 2

Calculation of the Mass of Aluminum Oxide

To calculate the mass of aluminum oxide that reacted with sulfuric acid, we need to consider the mass of the resulting salt and the percentage yield of the reaction.

Given: - Mass of the salt (aluminum sulfate) = 2.565 g - Percentage yield = 75% (75% of the theoretical yield was obtained)

To find the mass of the aluminum oxide that reacted, we can use the following formula:

Mass of aluminum oxide = (Mass of salt / Percentage yield) * 100

Substituting the given values:

Mass of aluminum oxide = (2.565 g / 75%) * 100

Calculating the result:

Mass of aluminum oxide = (2.565 g / 0.75) * 100 = 342 g

Therefore, the mass of aluminum oxide that reacted with sulfuric acid is 342 g.