A) 1 : 4
B) 4 : 1
C) 16 : 1
D) 2 : 1
Correct Answer: B
Solution :
Let the mass of \[{{H}_{2}}\] gas be \[xg\] and mass of \[{{O}_{2}}\] gas 4xg |
Molar \[{{\text{H}}_{\text{2}}}\] : \[{{O}_{3}}\] |
Mass 2 : 32 |
i.e. 1 : 16 |
\[\therefore \] Molar ratio = |
\[\frac{{{n}_{{{H}_{2}}}}}{{{n}_{{{O}_{2}}}}}=\frac{x/2}{4x/32}=\frac{x\times32}{2\times 4x}=\frac{4}{1}=4:1\] |
You need to login to perform this action.
You will be redirected in
3 sec