Question

How many grams of water can be prepared from the reaction of 26.0 g of MnO2 and 30.0 g oh HCl according to the following chemical equation?

I tried solving it but each time i wouldnt get an answer that was in the options

17, How many grams of water can be prepared from the reaction of 26.0 g 30.0 g of HCl according to the following chemical equation? MnO2 + 4HCI → MnCl2 + Cl2 + 2H20 A) 0.823 g B) 7.41 g C) 14.8 g D) 10.8g E) 58.4i

Answer 1

Given reactioon is

MnO₂ + 4 HCl ---> MnCl₂ + Cl₂ + 2 H₂O

Mass of MnO2 = 26 g

Molar mass of MnO2 = 86.94 g/mol

No. of moles of MnO2 = 26 g / 86.94 g/mol = 0.299 moles

Mass of HCl = 30 g

Molar mass of HCl = 36.46 g/mol

No. of moles HCl = 30 g / 36.46 g/mol = 0.823 moles

According to reaction stoichiometry

1 moles of MnO2 require 4 moles of HCl

0.299 moles of MnO2 require 4 * 0.299 moles = 1.196 moles but we have 0.823 moles of HCl only

so HCl is limiting reactant

According to reaction stoichiometry

4 moles of HCl will produce 2 moles of H2O

so 0.823 moles of HCl will produce 0.823 / 2 = 0.4115 moles of H2O

Molar mass of H2O = 18 g/mol

Mass of H2O = No. of moles * Molar mass = 0.4115 moles * 18 g/mol = 7.41 g

Answer is (B) 7.41 g