Question

0.5 gm of an organic compound with 60%. Carbon produce ______ gm of \(CO_{2}\) upon complete combustion

Answer 1

Correct Answer: 1.1

The correct answer is 1.1
Moles of Carbon = \(\frac{(0.5 \times 0.6)}{12}\)
Moles of \(CO_{2}\) = \(\frac{(0.5 \times 0.6)}{12}\)
Moles of \(CO_{2}\) = \(\frac{(0.5 \times 0.6)}{12}\times 44\)
Moles of \(CO_{2}\) = \(1.1 gm\)

Answer 2

To solve this problem, we need to use the stoichiometry of the combustion reaction between the organic compound and oxygen. The balanced chemical equation for the combustion of organic compounds with carbon and hydrogen can be written as:
CₓHᵧ + (\(x + \frac{y}{4}\)) O₂ → x CO₂ + \(\frac{y}{2}\) H₂O
where CₓHᵧ is the organic compound, x and y are the coefficients for carbon and hydrogen, respectively.
Since the organic compound contains 60% carbon, we can assume that it is a hydrocarbon with only carbon and hydrogen atoms. Therefore, we can write:
The molar mass of the organic compound = 12x + 1y \(\frac{grams}{mol}\)
The mass of carbon in the organic compound = 0.5 gm x 60% = 0.3 gm
The number of moles of carbon in the organic compound = 0.3 gm ÷ 12 \(\frac{grams}{mol}\) = 0.025 mol
According to the balanced equation, 1 mole of carbon produces 1 mole of CO₂ upon complete combustion
Therefore, 0.025 mol of carbon produces 0.025 mol of CO₂
The molar mass of CO₂ is 12 + 2x16 = 44 \(\frac{grams}{mol}\)
Therefore, the mass of CO₂ produced = 0.025 mol x 44 \(\frac{grams}{mol}\) = 1.1 gm
Therefore, 0.5 gm of the organic compound with 60% carbon produces 1.1 gm of CO₂ upon complete combustion.
Answer. 1.10