# Chemistry unit 5 MCQWatch

#1
Sulfur dioxide reacts with hydrogen sulfide to form water and sulfur. By considering
the changes in the oxidation numbers of sulfur, it can be deduced that, in this
reaction

A 1 mol of sulfur dioxide oxidizes 2 mol of hydrogen sulfide.
B 1 mol of sulfur dioxide reduces 2 mol of hydrogen sulfide.
C 2 mol of sulfur dioxide oxidizes 1 mol of hydrogen sulfide.
D 2 mol of sulfur dioxide reduces 1 mol of hydrogen sulfide.

How to find the ans?
0

3 weeks ago
#2
Just write out the equation
You'll get A
Also you should've posted this in the chemistry forum
0
#3
can you pls write the equations down I am very weak in chemistry

(Original post by Presto)
Just write out the equation
You'll get A
Also you should've posted this in the chemistry forum
0
3 weeks ago
#4
(Original post by irfan kobber)
can you pls write the equations down i am very weak in chemistry
Like this:
SO2 + 2H2S -> 2H2O + 3S
Last edited by Presto; 3 weeks ago
1
#5
by balancing the equation i understand that the answer is A but how do i use the changes in oxidation number of sulfur to find this(mentioned in the question)?
0
3 weeks ago
#6
(Original post by Irfan Kobber)
by balancing the equation i understand that the answer is A but how do i use the changes in oxidation number of sulfur to find this(mentioned in the question)?
In SO2 because each O has a state of -2, the total oxidation state is x + (-4) = 0 because there is no overall charge.
Thus x = 4.

As H2S it is -2 as each H is +1

In S, because its an element, the oxidation state is by default 0.

Therefore, the change for sulfur is
4 and -2 -> 0
This is 1SO2 and 1 H2S
To balance
1SO2 and 2H2S give
4 + (-4) -> 0
Which works

And because the S in H2S goes from -2 to 0 it is being oxidised (by the SO2)

Hence, A
2
X

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