if 5.0 g of each reactant were used for the following process, the limiting reactant would be:
2KMnO4 + 5Hg2Cl2 + 16HCl»
10HgCl2 + 2MnCl2 + 2KCl + 8H2O
a) KMnO4
b) HCl
c) H2O
d) Hg2Cl2
Answers
If 5g of each reactant were used for the following process the limiting reactant would be ..
2KMnO₄ + 5Hg₂Cl₂ + 16 HCl ⇒ 10 HgCl₂ + MnCl₂ + 2KCl + 8H₂O
solution : limiting reactant is the reactant which present insuficient in the reaction to proceed it further.
we should first find the number of moles of each reactant.
no of moles of KMnO₄ = given mass/mol wt of KMnO₄
= 5/158 = 0.0316 mol
no of moles of Hg₂Cl₂ = given mass/mol wt of Hg₂Cl₂
= 5/472 = 0.0106 mol
no of mole of HCl = given mass/mol wt of HCl
= 5/36.5= 0.137 mol
from chemical reaction, 2 moles of KMnO₄ reacts with 5 moles of Hg₂Cl₂ and 16 moles of HCl.
∴ 0.0316 mol of KMnO₄ reacts with 2.5 × 0.0316 = 0.079 moles of Hg₂Cl₂ [ but given only 0.0106 mol ]
∴ 0.0316 mol of KMnO₄ reacts with 8 × 0.0.316 = 0.2528 moles of HCl [ there is enough amount of HCl for the reaction. ]
therefore the limiting reactant is Hg₂Cl₂.