Question

How many grams go propane can be burned by 24g of O2?

Answer 1

How many grams go propane can be burned by 24g of $O_2$?

Solution:

Propane burning can occur as follows:

Complete oxidation of propane

1. $CH_{2}(CH_{3})_{2} + 5O_{2} = 3CO_{2} + 4H_{2}O$

Partial oxidation of propane

2. $2CH_{2}(CH_{3})_{2} + 7O_{2} = 3CO + 8H_{2}O$

3. $CH_{2}(CH_{3})_{2} + 2O_{2} = 3C + 4H_{2}O$

Find number of moles of oxygen

$$n(O_2) = m(O_2) / M(O_2)$$ $$n(O_2) = 24g / 32g \cdot mol^{-1} = 0.75mol$$

Find number of moles of propane $(n_1, n_2, n_3)$ for each reaction equation

$$\begin{array}{l} n_1(CH_2(CH_3)_2) = n(O_2) / 5 \\ n_1(CH_2(CH_3)_2) = 0.75mol / 5 = 0.15mol \\ \end{array}$$ $$\begin{array}{l} n_2(CH_2(CH_3)_2) = n(O_2) / 3.5 \\ n_2(CH_2(CH_3)_2) = 0.75mol / 3.5 = 0.214mol \\ \end{array}$$ $$\begin{array}{l} n_3(CH_2(CH_3)_2) = n(O_2) / 2 \\ n_3(CH_2(CH_3)_2) = 0.75mol / 2 = 0.375mol \\ \end{array}$$

Find weight of propane for each reaction equation

$$\begin{array}{l} m_1(CH_2(CH_3)_2) = n_1(CH_2(CH_3)_2) * M(CH_2(CH_3)_2) \\ m_1(CH_2(CH_3)_2) = 0.15mol * 44g \cdot mol^{-1} = 6.6g \\ \end{array}$$ $$\begin{array}{l} m_2(CH_2(CH_3)_2) = n_2(CH_2(CH_3)_2) * M(CH_2(CH_3)_2) \\ m_2(CH_2(CH_3)_2) = 0.214mol * 44g \cdot mol^{-1} = 9.416g \\ \end{array}$$ $$\begin{array}{l} m_3(CH_2(CH_3)_2) = n_3(CH_2(CH_3)_2) * M(CH_2(CH_3)_2) \\ m_3(CH_2(CH_3)_2) = 0.375mol * 44g \cdot mol^{-1} = 16.5g \\ \end{array}$$

Answer: In 24 g of oxygen during the partial oxidation 16.5 g of propane can be burnt.