How many grams of aluminum do you need to take to get 170 grams of aluminum sulfate?

We carry out the solution:
1. We write down the equation:
2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2 – OBP, aluminum sulfate was obtained;
2. We calculate the molar masses of substances:
M (Al) = 26.98 g / mol;
M Al2 (SO4) 3 = 341.96 g / mol.
3. Determine the amount of mol of aluminum and metal salt – Al:
Y Al2 (SO4) 3 = m / M = 170 / 341.96 = 0.5v = mol;
X mol (Al) – 0.5 mol Al2 (SO4) 3;
-2 mol – 1 mol from here, X mol (Al) = 0.5 * 2 = 1 mol.
4. Find the mass of the metal:
m (Al) = Y * M = 1 * 26.98 = 26.98 g.
Answer: 26.98 g of aluminum metal is needed to carry out the reaction.