Question

1. If you have 10.0g of each reactant, determine the limiting reagent and the amount of H₂O produced.

MnO₂ (s) + H₂SO₄ (l) ---> Mn(SO₄)₂ (s) + H₂O (l)

Can you go over each step?

Answer 1

MnO₂ (s) + 2 H₂SO₄ (l) ---> Mn(SO₄)₂ (s) + 2 H₂O (l)

1 mole       2 mole

no of moles of MnO2 = Weight of MnO2/Gram molar mass of MnO2

                                    = 10/87   = 0.1149 moles

no of moles of H2SO4 = Weight of H2SO4/Gram molar mass of H2SO4

                                       = 10/98 = 0.102 moles

from balanced equation

2 moles of H2SO4 react with 1 mole of MnO2

0.102 moles of H2SO4 react with = 1*0.102/2 = 0.051 moles of MnO2

Mno2 is excess reagent

limiting reagent is H2SO4

2 moles of H2So4 react mno2 to form 2 moles of H2O

0.102 moles of H2So4 react with MnO4 to form = 2*0.102/2 = 0.102 moles of H2O

mass of H2O = no of moles of H2O * Gram molar mass

                     = 0.102*18 =1.836gm of H2O >>>> answer