Question

given that 16.0g of MnO2 and 30,0g of HCl react according to: MnO2 + 4HCl ---> MnCl2 + Cl2 + 2H20 a: what is the limiting reagent? b: what mass of MnCl2 could be produced? c: how much of each reagent remains when the reaction is complete? d. assuming a % yeild of 71.2%, caculate the auctual yeild of McCl2

Answer 1

a) molar mass of MnO₂ = 86.9368 gm/mole

molar mass of HCl = 36.46094 gm/mol then 4 mole of HCl = 36.46094 4 = 145.84376 gm

According to reaction 1 mole of MnO₂ react with 4 mole of HCl that mean 86.9368 gm MnO₂ react with 145.84376 gm of HCl then to react with 16 gm of MnO₂ required HCl = 145.84376 16 / 86.84376 = 26.87 gm

but HCl given 30 gm therefore HCl is excess ragent and MnO₂ is limiting reagent

b) molar mass of MnCl₂ = 125.844 gm/mol

According to reaction 1 mole of MnO₂ produce 1 mole of MnCl₂ that mean 86.84376 gm of MnO₂ produce 125.844 gm of MnCl₂ then 16 gm MnO₂ produce 125.844 16 /86.84376 = 23.185 gm of MnCl₂ produced.

c)

according to reaction 1 mole of MnO₂ react with 4 mole of HCl that mean 86.9368 gm MnO₂ react with 145.84376 gm of HCl then to react with 16 gm of MnO₂ required HCl = 145.84376 16 / 86.84376 = 26.87 gm

but HCl given 30 gm therefore HCl is excess ragent and MnO₂ is limiting reagent

HCl remain after complete reaction = 30.0 - 26.87 = 3.13 gm

d)23.185 gm of MnCl₂ = 100% yield then 71.2% = 71.2 23.185 / 100 = 16.51 gm is actual yield