Question

Технический алюминий массой 5.9 г обработали избытком соляной кислоты при этом выделилось 6.72 л водорода . определите массовую долю примесей в техническом образце металла

Answer 1

Дано:
m (Al тех) = 5.9г
V (H2) = 6.72л

Найти:
W (Al) - ?

Решение:
2Al + 6HCl = 2AlCl3 + 3H2

n (H2) = V/Vm = 6.72 / 22.4 = 0.3 моль

n (Al) = 2/3 n (H2) = 2/3 * 0.3 = 0.2 моль

m (Al) = n * M = 0.2 * 27 = 5.4г

m (примесей) = m (Al тех) - m (Al) = 5.9 - 5.4 = 0.5г

W (примесей) = m (примесей) / m (Al тех) = 0.5 / 5.9 = 0.0847 = 8.47%

Ответ: W (примесей) = 8.47%

Answer 2

Calculation of Impurity Mass Fraction in Technical Aluminum

To determine the mass fraction of impurities in the technical aluminum sample, we need to use the information provided about the reaction between the aluminum and excess hydrochloric acid.

According to the given information: - Mass of technical aluminum = 5.9 g - Volume of hydrogen gas evolved = 6.72 L

To calculate the mass fraction of impurities, we need to find the mass of pure aluminum in the sample and subtract it from the total