Question

Example 32
How many gram of oxygen (O2) is required to completely react with 0.200 g of hydrogen (H2) to
yield water (H2O).? Also calculate the amount of water formed. (molecular mass H = 2, O = 32)
​

Answer 1

Given mass of Hydrogen = 0.200 grams

Molecular mass of Hydrogen = 2 units.

Thus,

Number of moles of hydrogen in 0.200 gram of it will be :-

$= \frac{0.200}{2} \\ \\ \\ = 0.1$

Now, we have the reaction :-

$2H_2 +O_2 =2H_2 O$

Looking at molar coefficients, we can say that :-

2 moles of Hydrogen reacts with 1 mole of Hydrogen to produce 2 moles of Water.

If we have 0.1 mole of Hydrogen :-

Required oxygen = 0.05 moles

Formed product (water) will be 0.1 moles.

From this :-

Mass of Oxygen required =

$= 0.05 \times 32 \\ \\ \\ = 1.6$

Thus :-

1.6 gram of oxygen is required.

Now,

Number of moles of water formed =0.1

Molecular mass of water = 18 units

Amount of water formed :-

$18 \times 0.1 \\ \\ \\ = 1.8 \: grams$

$\rule{200}{2}$

Answer 2

Explanation:

2H2+02 ---->>>> 2H20

2gm produce 32 g of 02

so 0.200g produce 32×0.200/2

= 6.400/2

= 3.2 gm