Al + HCl = ? уравнение реакции Какие продукты образуются в результате взаимодействия Al + HCl = ? Надо записать молекулярное уравнение реакции, а также решить задачу. Вот условие: какая масса соли образуется при растворении 10 г алюминия в 150 г 5%-ного раствора соляной кислоты? Какое из веществ будет в избытке?

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Molecular equation:
2Al + 6HCl -> 2AlCl3 + 3H2

To find the mass of salt formed when 10 g of aluminum reacts with 150 g of 5% hydrochloric acid solution, we need to first calculate the amount of hydrochloric acid present:

150 g * (5/100) = 7.5 g of HCl

Now, we need to determine the limiting reactant.

Molar mass of Al = 26.98 g/mol
Molar mass of HCl = 36.46 g/mol

10 g of Al is equivalent to 0.370 moles of Al.
7.5 g of HCl is equivalent to 0.206 moles of HCl.

From the balanced equation, it can be seen that 2 moles of Al reacts with 6 moles of HCl to form 2 moles of AlCl3.

Since 2 moles of AlCl3 are formed from 2 moles of Al, we can say that 0.370 moles of Al will produce 0.370 moles of AlCl3.

The molar mass of AlCl3 is 133.34 g/mol.

0.370 moles * 133.34 g/mol = 49.34 g of AlCl3 formed.

Therefore, the mass of salt (AlCl3) formed when 10 g of aluminum reacts with 150 g of 5% HCl solution is 49.34 g.

HCl is in excess in this reaction.