Law of Definite Proportion. When 60.0 g of carbon is burned in 160.0 g of oxygen, 220.0 g of carbon dioxide is formed. What mass of carbon dioxide is formed when 60.0 g of carbon is burned in 750.0 g of oxygen? Explain and support your answer by calculating for the number of moles involved for each of the above reactions. (5 points)

The Law of Definite Proportions, also known as Proust's Law, states that a chemical compound will always have exactly the same proportion of elements by mass. This means that the ratio of the mass of carbon to the mass of oxygen in carbon dioxide is always the same.

In the first reaction, 60.0 g of carbon reacts with 160.0 g of oxygen to form 220.0 g of carbon dioxide.

First, let's calculate the number of moles involved in this reaction.

The molar mass of carbon (C) is approximately 12.01 g/mol and the molar mass of oxygen (O2) is approximately 32.00 g/mol.

For carbon: Number of moles = mass / molar mass = 60.0 g / 12.01 g/mol = 4.996 moles

For oxygen: Number of moles = mass / molar mass = 160.0 g / 32.00 g/mol = 5.000 moles

This means that the ratio of moles of carbon to moles of oxygen in this reaction is approximately 1:1, which is consistent with the chemical formula for carbon dioxide (CO2).

In the second reaction, we are asked to find out what mass of carbon dioxide is formed when 60.0 g of carbon is burned in 750.0 g of oxygen.

Since the ratio of moles of carbon to moles of oxygen is 1:1, we can say that 4.996 moles of carbon will react with 4.996 moles of oxygen to form 4.996 moles of carbon dioxide.

The molar mass of carbon dioxide (CO2) is approximately 44.01 g/mol.

Mass of carbon dioxide = number of moles * molar mass = 4.996 moles * 44.01 g/mol = 219.9 g

Therefore, when 60.0 g of carbon is burned in 750.0 g of oxygen, approximately 219.9 g of carbon dioxide is formed. This is consistent with the Law of Definite Proportions.

The Law of Definite Proportions, also known as Proust's Law, states that a chemical compound always contains exactly the same proportion of elements by mass. This means that the ratio of the mass of carbon to the mass of oxygen in carbon dioxide is always the same.

In the first reaction, 60.0 g of carbon reacts with 160.0 g of oxygen to produce 220.0 g of carbon dioxide.

First, let's calculate the number of moles involved in this reaction.

The molar mass of carbon (C) is approximately 12.01 g/mol and the molar mass of oxygen (O2) is approximately 32.00 g/mol.

Number of moles of carbon = 60.0 g / 12.01 g/mol = 4.996 moles Number of moles of oxygen = 160.0 g / 32.00 g/mol = 5.000 moles

The molar mass of carbon dioxide (CO2) is approximately 44.01 g/mol.

Number of moles of carbon dioxide = 220.0 g / 44.01 g/mol = 4.998 moles

This shows that the reaction is approximately: C + O2 -> CO2

In the second reaction, we have the same amount of carbon (60.0 g), but more oxygen (750.0 g). However, according to the Law of Definite Proportions, the ratio of carbon to oxygen in carbon dioxide is always the same. Therefore, the amount of carbon dioxide formed will be the same as in the first reaction, regardless of the excess oxygen.

So, when 60.0 g of carbon is burned in 750.0 g of oxygen, 220.0 g of carbon dioxide is formed.