Answer
Nitric Oxide (NO) which is produced by internal combustion engines. It oxidizes in air to form Nitrogen Dioxide ($NO_{2})$.
i) What Mass of O combines with 3.0 g of N in NO.
? g of O = 3.0 g N $\times$[16.0 g O/14.0 g N) =3.43 g O
Mass of Oxygen required = 3.43 gram.
ii) What Mass of O combines with 3.0 g of N in ($NO_{2})$.
? g of O = 3.0 g N $\times$[32.0 g O/14.0 g N) =6.86 g O
Mass of Oxygen required = 6.86 gram.
iii) Taking the ratio :
g of O in NO /g of O in ($NO_{2})$ = 3.43 /6.86 = 1/2
This follows law of Multiple proportions which states that When two elements, X and Y form more than one compound, the ratio of the masses of element
Y that combine with a given mass of element X in each of the compounds can be expressed by small whole numbers.
Work Step by Step
Nitric Oxide (NO) which is produced by internal combustion engines. It oxidizes in air to form Nitrogen Dioxide ($NO_{2})$.
Molar mass of Nitrogen = 14.0 g
Molar mass of Oxygen = 16.0 g
i) What Mass of O combines with 3.0 g of N in NO.
In Nitric oxide 14.0 g of nitrogen combines with 16.0 g of oxygen. Then we can calculate 3.0 g of nitrogen combine with how much amount of Oxygen.
? g of O = 3.0 g N $\times$[16.0 g O/14.0 g N) =3.43 g O
Mass of Oxygen required = 3.43 gram.
ii) What Mass of O combines with 3.0 g of N in ($NO_{2})$.
In Nitrogen dioxide 14.0 g of nitrogen combines with 32.0 g of oxygen. Then we can calculate 3.0 g of nitrogen combine with how much amount of Oxygen.
? g of O = 3.0 g N $\times$[32.0 g O/14.0 g N) =6.86 g O
Mass of Oxygen required = 6.86 gram.
iii) Taking the ratio :
g of O in NO /g of O in ($NO_{2})$ = 3.43 /6.86 = 1/2
This follows law of Multiple proportions which states that When two elements, X and Y form more than one compound, the ratio of the masses of element
Y that combine with a given mass of element X in each of the compounds can be expressed by small whole numbers.
