Question

A certain mass of carbon reacts with 23.3 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that same mass of carbon to form carbon dioxide, according to the law of multiple proportions.

Answer 1

Answer: 46.6 g of Oxygen

Law of multiple proportions, states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

To form Carbon monoxide ( CO ), 12 parts by mass of carbon combines with 16 parts by mass of oxygen. To form Carbon dioxide ( CO2 ),12 parts by mass of carbon combines with 32 parts by mass of oxygen.

By applying Law of multiple proportions we get that,

Ratio of the masses of oxygen that combines with a fixed mass of carbon (12 parts) is equal to 16:32 or 1:2.

As it is given that 23.3 g of Oxygen reacts with a fixed mass of Carbon to produce CO,by applying the Law of multiple proportions we arrive at the solution that 46.6 g of Oxygen reacts with the same fixed mass of C to produce CO2.

Answer 2

Answer: The mass of oxygen than would react in $CO_2$ is 46.6 grams

Explanation:

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number.

We are given:

Mass of oxygen in carbon monoxide = 23.3 grams

The chemical formula of carbon monoxide is CO

We know that:

Molar mass of carbon = 12 g/mol

Molar mass of oxygen = 16 g/mol

In CO molecule:

If 16 g of oxygen element reacts with 12 g of carbon

o, 23.3 g of oxygen element will react with = $\frac{12}{16}\times 23.3=17.475g$ of carbon

In $CO_2$ molecule:

Taking mass of carbon = 17.475 grams

If 12 g of carbon reacts with 32 g of oxygen

So, 17.475 g of carbon will react with = $\frac{32}{12}\times 17.475=46.6g$ of oxygen

Hence, the mass of oxygen than would react in $CO_2$ is 46.6 grams