Answer :
MnO2 + 4HCl = MnCl2 + Cl2 + 2H2O
m(HCl) = 36.45/48.2 = 0.76
every 4 moles of HCl will only react with 1 mole MnO2 and there is less HCl therefore HCl will be used up first.
Excess MnO2 = 0.86 - (0.76/4) = 0.67m
grams MnO2 = 0.67x 86.93= 58.24grams left over.
Every 4 moles HCl will only form 1 mole Cl2 so m(Cl2)=0.76/4=0.19
0.19x70.9=13.47grams Chlorine formed
m(HCl) = 36.45/48.2 = 0.76
every 4 moles of HCl will only react with 1 mole MnO2 and there is less HCl therefore HCl will be used up first.
Excess MnO2 = 0.86 - (0.76/4) = 0.67m
grams MnO2 = 0.67x 86.93= 58.24grams left over.
Every 4 moles HCl will only form 1 mole Cl2 so m(Cl2)=0.76/4=0.19
0.19x70.9=13.47grams Chlorine formed
Answer:
Amount of Cl2 produced = 23 g
Amount of excess reagent MnO2 remaining = 46 g
Explanation:
The reaction between MnO2 and HCl forming MnCl2, Cl2 and H2O can be expressed as:
[tex]4HCl + MnO2 \rightarrow Cl2 + MnCl2 + 2H2O[/tex]
The amount of Cl2 produced will be determined by the amount of the limiting reactant.
Moles of MnO2 = 0.86 moles
[tex]Moles\ HCl = \frac{Mass\ HCl}{Mol.wt.\ HCl} = \frac{48.2g}{36.46g/mol} =1.32moles[/tex]
The mole ratio between HCl:MnO2 = 4:1
If all 1.32 moles of HCl were used up then based on the mole ratio, the moles of MnO2 that would be needed is:
[tex]Moles\ MnO2\ needed = \frac{1}{4} *moles\ of\ HCl\\\\= \frac{1}{4} *1.32 = 0.33 moles[/tex]
However, there are 0.86 moles of MnO2 i.e. there is an excess of MnO2. Hence HCl is the limiting reactant.
Based on the reaction stoichiometry:
4 moles of HCl produces 1 mole of Cl2
[tex]Moles\ Cl2\ produced = \frac{1}{4} *moles\ of\ HCl\\\\= \frac{1}{4} *1.32 = 0.33 moles[/tex]
[tex]Mass\ Cl2 = moles*mol wt = 0.33moles*70.9g/mol = 23 g[/tex]
Excess reagent = MnO2
Moles of MnO2 remaining = 0.86 -0.33=0.53 moles
[tex]Mass\ MnO2\ remaining = moles*mol wt = 0.53moles*86.9g/mol = 46g[/tex]