1 kg of an alloy containing 24% magnesium and 765 aluminum was dissolved in 35.38% hydrochloric

1 kg of an alloy containing 24% magnesium and 765 aluminum was dissolved in 35.38% hydrochloric acid solution (ϸ = 1.18). Please provide the correct statement. A) the total mass of the salts that are formed is 3758 g. B) the volume of the hydrochloric acid solution required to dissolve the alloy is 12 liters. C) the mass of the obtained magnesium chloride is 1 kg. D) there is no correct statement.

1) write the reaction equations:
Mg + 2 HCl => MgCl2 + H2
2 Al + 6 HCl => 2 AlCl3 + 3 H2

2) find the mass of Mg in the alloy:
m (Mg) = ω (Mg) * m (alloy) / 100% = 24% * 1000/100% = 240 g

3) find the amount of substance Mg and Al:
n (Mg) = m (Mg) / Mr (Mg) = 240/24 = 10 mol
n (Al) = m (Al) / Mr (Al) = 765/27 = 28.33 mol

4) find the amount of substance MgCl2 and AlCl3 (according to the reaction equations):
n (MgCl2) = n (Mg) = 10 mol
n (AlCl3) = n (Al) = 28.33 mol

5) find the mass of the substance MgCl2 and AlCl3:
m (MgCl2) = n (MgCl2) * Mr (MgCl2) = 10 * 95 = 950 g
m (AlCl3) = n (AlCl3) * Mr (AlCl3) = 28.33 * 133.5 = 3778.06 g
m (MgCl2) + m (AlCl3) = 4728.06 g

A) and C) are incorrect

6) find the amount of HCl substance (according to the reaction equations):
n (HCl) = 2 * n (Mg) = 2 * 10 = 20 mol
n (HCl) = 3 * n (Al) = 3 * 28.33 = 84.99 mol
n (HCl) = 20 + 84.99 = 104.99 mol

7) find the mass of the substance HCl:
m (HCl) = n (HCl) * Mr (HCl) = 104.99 * 36.5 = 3832.14 g

8) find the mass of the HCl solution:
m solution (HCl) = m (HCl) * 100% / ω (HCl) = 3832.14 * 100% / 35.38% = 10831.37 ml

9) find the volume of the HCl solution:
V solution (HCl) = m solution (HCl) / p (HCl) = 10831.37 / 1.18 = 9179.13 ml = 9.18 L

Statement B) is incorrect

10) True statement – D)