Answer:
The answer to your question is:
a) 1.12 l of Cl2
b) 6.3 g of MnCl2
Explanation:
Reaction
4HCl + MnO2 ⇒ Cl2 + 2H2O + MnCl2
Data
HCl 5M, 50 ml (0.05 l)
MnO2 = 5 g
MW MnO2 = 87 g
MW MnCl2 = 126 g
Process
For HCl
M = n/V
n = MV
n = 5(0.05) = 0.25 moles from the reaction 4(0.05) = 1 mol of HCl
For MnO2
87 g --------------------- 1 mol
5 g -------------------- x
x = 5/87 = 0.05 mol of MnO2
The proportion in the reaction HCl to MnO2 is 4:1 but from the operations,
1: 0.05, the from this we see that MnO2 is the limiting reactant, and we take MnO2 to obtain the values of the questions.
a) From the reaction 1 mol of MnO2 --------------- 1 mol of Cl2
0.05 mol --------------- x
x = 0.05 moles of Cl2
1 mol of Cl2 --------------- 22.4 l
0.05 moles ------------- x
x = 0.05 x 22.4/1 = 1.12 l
b) From the reaction 1 mol of MnO2 ----------------- 1 mol of MnCl2
0.05 mol of MnO2 ------------- x
x = 0.05 moles of MnCl2
1 mol of MnCl2 ---------------- 126 g of MnCl2
0.05 mol of MnCl2 ------------ x
x = 0.05 x 126 = 6.3 g
