Question

What is the volume required of a 20.0% HCl solution of density 1.20 g/ml to prepare 363.0 g of AsCl3​ according to the equations? (As=75,Cl=35.5) 2KMnO+16 HCl→2KCl+2MnCl2​+5Cl2​+8H2​O 2As+3Cl2​→2AsCl3​
A 2.56 l
B 0.73 l
C 1.46 l
D 2.92 l​

Answer 1

Correct option is
A
2.56 l
mole of AsCl
3
​
=
75+3(35.5)
36.6
​
=2 mole
2As+3Cl
2
​
→2AsCl
3
​

2 mole $$=
moleCl_2$$
2 KMnO
4
​
+18HCl→2KCl+2MnCl
2
​
+5Cl
2
​
+8H
2
​
O
K HCl=5Cl
2
​

λH=3Cl
2
​

λ=
5
3x16
​
9.6 mole
volume of =
1.20 g/ml
100 g
​
=83.33 ml
=0.08333 L
mole of HCl in solution =
36.5 g/ml
0