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Limiting reagent
Concepts tested here
- Excess reagent
- Limiting reagent
All Questions
2014 AIPMT 2 questions
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When 22.4 litres of $H_2(g)$ is mixed with 11.2 litres of $Cl_2(g)$, each at S.T.P., the moles of HCl(g) formed is equal to:$n_{H_2} = \frac{22.4}{22.4} = 1$ mol; $n_{Cl_2} = \frac{11.2}{22.4} = 0.5$ mol
$H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$
$Cl_2$ is the limiting reagent: 0.5 mol $Cl_2$ gives $0.5\times2 = 1$ mol HCl (0.5 mol $H_2$ is left). -
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)$n_{Mg} = \frac{1}{24} = 0.0416$ 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
