Mg(s) + 2 HCl(aq)  MgCl2 (aq) + H2(g) b. What mass of magnesium will react with 8.5 liters of 0.7 M hydrochloric acid?

No. of moles of HCl reacted = (0.7 mol/L) × (8.5 L) = 5.95 mol

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mole ratio Mg : HCl = 1 : 2
No. of moles of Mg reacted = (5.95 mol) × (1/2) = 2.975 mol

Molar mass of Mg = 24.3 g/mol
Mass of Mg reacted = (2.975 mol) × (24.3 g/mol) = 72.3 g

you have 0.7 moles of HCL per 1000grams of solution or 1 LiTER, multiply 8.5 time 0.7 to get total moles of HCL so the stoichiometric balance of your equation indicgs 1 mole Mg per 2 moles HCL so......now you have 8.5X0.7/2 to get the total moles of Mg whose mwt is 24 give or take a few quarks, so now multiply the previous computation by that number and VIOLA, il est finis.

moles(HCl) = 0.7M x 8500 mL / 1000
moles(HCl) = 5.95 moles. (THis equivalent to '2' in the molar ratios).

moles(Mg) = 5.95 / 2 = 2.975 moles (This is equivalent to '1' in the molar ratios).

moles = mass(g) / Ar
mass(g) = moles x Ar
mass(g) = 2.975 x 24,3 = 72.2925 g

...
M = n / V
0.7 = n / 8.5
n = 5.95

. . 145 g {answer]
. . Mg(s) . . + . . 2 HCl(aq) . . → . . MgCl2 (aq) . . + . . H2(g)
(24.3g/mol)
. . 5.95mol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.95mol
. . . .|______________1 to 1_______________________|

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