Question

Assuming that the results given below taken together illustrate the law of reciprocal proportions . Calculate the masses of oxygen and hydrogen in 1.25 g of water. 1.) 0.46 g of magnesium produces 0.77 g magnesium oxide. 2.) 0.82 g of magnesium liberates 760 ml of hydrogen at STP from an acid . [ Weight of 1 ml of Hydrogen at STP = 0.00009 g]

Answer 1

Given:

Assuming that the results given below taken together illustrate the law of reciprocal proportions .

To find:

Calculate the masses of oxygen and hydrogen in 1.25 g of water.

1.) 0.46 g of magnesium produces 0.77 g magnesium oxide.

2.) 0.82 g of magnesium liberates 760 ml of hydrogen at STP from an acid .

Solution:

From given, we have,

1) 0.46 g of magnesium produces 0.77 g magnesium oxide.

Mg = 0.46 g

O= 0.77 - 0.46 = 0.31 g

Therefore, the weight of oxygen that combines with 1 g of Mg,

Mg = 0.31/0.46 = 0.674 g

2) 0.82 g of magnesium liberates 760 ml of hydrogen at STP from an acid .

H2 = 0.00009 × 760 = 0.068 g

Thus, the weight of H2 that will liberated by 1 g of Mg,

Mg = 0.068/0.82 = 0.083 g

According to (1) and (2) the ratio of the weight of O:H that combine-with/displaced-by the same weight of magnesium is,

0.674/0.083 = 8:1

∴ O:H = 8:1