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1.0 g of magnesium is burnt with 0.56 g $O_2$ in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24; O = 16)
A
Mg, 0.16 g
B
$O_2$, 0.16 g
C
Mg, 0.44 g
D
$O_2$, 0.28 g
Detailed Solution
$n_{Mg} = \frac{1}{24} = 0.0416$ mol; $n_{O_2} = \frac{0.56}{32} = 0.0175$ mol
$Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)$
Mg used $= 2\times0.0175 = 0.035$ mol; Mg left $= 0.0416 - 0.035 = 0.0066$ mol
Mass of Mg left $= 0.0066\times24 = 0.16$ g
$Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)$
Mg used $= 2\times0.0175 = 0.035$ mol; Mg left $= 0.0416 - 0.035 = 0.0066$ mol
Mass of Mg left $= 0.0066\times24 = 0.16$ g