If 0.86 mole of MnO2 and 57.2 g of HCl react, how many grams of Cl2will be produ
ID: 689658 • Letter: I
Question
If 0.86 mole of MnO2 and 57.2 g of HCl react, how many grams of Cl2will be produced?MnO2 + 4HCl ---- > MnCl2 + Cl2+ 2H2O
___ g.
Thank you.
MnO2 + 4HCl ---- > MnCl2 + Cl2+ 2H2O
___ g.
Thank you.
Explanation / Answer
MnO2 + 4HCl ---- > MnCl2 + Cl2 + 2H2O here 1 mole of MnO2 = 55+ 16*2 = 87 g and 1 mole of HCl = 1+ 35.5 g = 36.5 g now 4 moles of HCl = 4*36.5 g = 146 g 1 mole of Cl2 = 71 g so 87 g of MnO2 reacts with 146 g of HCl now when we take, 0.86 mole of MnO2 and 57.2 g of HCl 57.2 g of HCl = 57.2/36.5 mole = 1.567 moles so 1 mole of MnO2 needs 4 moles of HCl so 0.86 mole of MnO2 needs = 4*0.86 moles = 3.44 mole of HCl but HCl given is less so here HCl is limiting reagent and all the calculation will bedone using it only. so 4 moles of HCl produce 1 mole of Cl2 so 1.567 mole of HCl produces = 0.39175 mole of Cl2 = 0.39175*71 g of Cl2 = 27.814 g of Cl2
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