Sagot :
Answer:
[tex]a. \ \ \ \text{The mass of ethylene is 11.375 g}. \\ \\ b. \ \ \ \text{The volume of water produced is 7.875 ml}.[/tex]
Explanation:
a. According to the chemical equation for the combustion of ethylene, 5
moles of oxygen are required to react with 2 moles of ethylene, hence
[tex]m_{C_{2}H_{2}} \ = \ 35 \ \text{g} \ \times \ \displaystyle\frac{1 \ \text{mol}}{\left(2 \times 16 \right) \text{g}} \ \times \ \displaystyle\frac{2 \ \text{mol}}{5 \ \text{mol}} \ \times \ \displaystyle\frac{\left(12 \ \times \ 2 \ + \ 1 \ \times \ 2 \right) \ \text{g}}{1 \ \text{mol}} \\ \\ m_{C_{2}H_{2}} \ = \ 11.375 \ \text{g}[/tex]
b. Similarly, 5 moles of oxygen are required as a reactant to produce 2
moles of water, thus
[tex]m_{H_{2}O} \ = \ 35 \ \text{g} \ \times \ \displaystyle\frac{1 \ \text{mol}}{\left(2 \times 16 \right) \text{g}} \ \times \ \displaystyle\frac{2 \ \text{mol}}{5 \ \text{mol}} \ \times \ \displaystyle\frac{\left(1 \ \times \ 2 \ + \ 16 \right) \ \text{g}}{1 \ \text{mol}} \\ \\ m_{H_{2}O} \ = \ 7.875 \ \text{g}[/tex]
Given that the density of water is [tex]1.0 \ \text{g ml}^{-1}[/tex], therefore
[tex]\rho_{H_{2}O} \ = \ \displaystyle\frac{m_{H_{2}O}}{V_{H_{2}O}} \\ \\ V_{H_{2}O} \ = \ \displaystyle\frac{m_{H_{2}O}}{\rho_{H_{2}O}} \\ \\ V_{H_{2}O} \ = \ \displaystyle\frac{7.875 \ \text{g}}{1.0 \ \text{g ml}^{-1}} \\ \\ V_{H_{2}O} \ = \ 7.875 \ \text{ml}[/tex]
11.375 g ethylene is consumed if 35.0 grams of oxygen is allowed to react and 7.875 ml of water will be produced when the 35.0 grams of oxygen is allowed to react.
What are moles?
A mole is defined as 6.02214076 ×[tex]10^{23}[/tex] of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
According to the chemical equation for the combustion of ethylene, 5
moles of oxygen are required to react with 2 moles of ethylene, hence
[tex]m_{_C_2_H_2}[/tex] [tex]= 35gX \frac{1 mol}{(2 \;X\; 16)g } X \frac{2 mol}{(5\;mol )} X \frac{26g}{1 mol }[/tex]
[tex]m_{_C_2_H_2} = 11.375 g[/tex]
Similarly, 5 moles of oxygen are required as a reactant to produce 2 moles of water, thus
[tex]m_{_H_2_O} = 35gX \frac{1 mol}{(2 \;X\; 16)g } X \frac{2 mol}{(5\;mol )} X \frac{18g}{1 mol }[/tex]
[tex]m_{_H_2_O} = 7.875 g[/tex]
Given that the density of water is [tex]1.0g ml^{-1}[/tex], therefore
[tex]Density = \frac{Mass \;of \;water}{Volume \;of \;water}[/tex]
[tex]Volume \;of \;water = \frac{7.875}{1.0 g ml^{-1}}[/tex]
The volume of water = 7.875 ml
Learn more about moles here:
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