Question

The volume of oxygen required at STP to form 36g of water according to the given reaction is:

H2(g)+1/2O2(g)->H2O(g)

a) 11.2L b) 2.4L c)5.6L

d)22.4L

Answer 1

Firstly, IF u write the formula on basis of their weights then :
   2*1 + 1/2*32 ===== 18
 => 18 ==== 18
Ratio of oxygen in 18g of water = 16/18
Therefore, amount of Oxygen in 36g of Water will be = 32g

Now, mole(amount of gas)= Weight in Gram / Molecular weight
                                              32/32
1 mole of oxygen

AT STP
 Volume of 1 mole of any gas is 22.4 L will be required.

Therefore, 22.4 L Oxygen is required.



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