Question

As per the law of constant proportion; carbon and oxygen combine in the ratio of 3:8. Compute the mass of oxygen that would be required to completely react with $6.9g$ of carbon.

Answer 1

Formulate the reaction for the combustion of carbon with oxygen and then calculate the weight based on the number of moles of reactants consumed.

Complete answer:
It has been given that by the law of constant proportion, carbon and oxygen combine in the proportion 3:8. The law of constant proportion states that in the reactants and products, all the elements are present in a specific proportion that does not change. This implies that the reactants will always combine in a specific proportion mass wise to give the product.
The question talks about the proportion 3:8; from this, we can derive the conclusion that every $3g$ of carbon has to combine with $8g$ of oxygen to complete the combustion reaction. The question is to find the mass of oxygen required to completely react with $6.9g$ of carbon. We can use the law of constant proportion and cross multiplication to find the mass of oxygen which we will denote by $x$. Thus, solving for $x$ we will get:

& \frac{3}{8}=\frac{6.9}{x} \\\ & x=\frac{6.9\times 8}{3} \\\ & x=18.4g \\\ \end{aligned}$$ Thus, the value of $x$ is $18.4g$. Hence, according to the law of constant proportions, we can say that $18.4g$ of oxygen is required to combine with $6.9g$ of carbon. **Additional information:** This is a combustion reaction where carbon undergoes combustion in the presence of oxygen to form carbon dioxide. The reaction is as follows: $$C+{{O}_{2}}\to C{{O}_{2}}$$ Here, we can see that one mole of each carbon and oxygen gives one mole of carbon dioxide. We can find the answer to this question by first converting the mass into moles too. **Note:** Remember that while referring to the law of constant proportion, we are always referring to mass since it remains constant throughout the reaction. The number of moles may not always add up to be the same on both sides. In this reaction itself, we have 2 moles on the LHS and 1 mole on the RHS. We can see that they are not adding up so they cannot be involved in the law of constant proportions.