Mg(s) + HCl (aq) ---> MgCl2(aq) + H2(g) Consider the above unbalanced equation. What volume of H2 is produced at 340 mm Hg and 73.4°C when 0.86 g of Mg reacts with excess HCl? Use molar masses with at least as many significant figures as the data given
Mg + 2 HCl —> MgCl2 + H2
Molar mass of Mg = 24.31 g/mol
mass of Mg = 0.86 g
mol of Mg = (mass)/(molar mass)
= 0.86/24.31
= 0.0354 mol
From balanced chemical reaction, we see that
when 1 mol of Mg reacts, 1 mol of H2 is formed
mol of H2 formed = moles of Mg
= 0.0354 mol
we have:
P = 340.0 mm Hg
= (340.0/760) atm
= 0.4474 atm
n = 0.0354 mol
T = 73.4 oC
= (73.4+273) K
= 346.4 K
we have below equation to be used:
P * V = n*R*T
0.4474 atm * V = 0.0354 mol* 0.0821 atm.L/mol.K * 346.4 K
V = 2.3 L
Answer: 2.3 L
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