Question

5. In an oxyacetylene welding torch, acetylene (C2H2) burns in pure oxygen with a very hot flame. The reaction is: 2 C2H2 + 5 O2 → 4 CO2 + 2 H2O.
a. What is the mass of C2H2 that is consumed if 35.0 grams of oxygen is allowed to react?


b. How many milliliters of water will be produced when the 35.0 grams of oxygen from part a. above is allowed to react. The density of water is 1.0 gram

Answer 1

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 ×$10^{23}$ 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

$m_{_C_2_H_2}$  $= 35gX \frac{1 mol}{(2 \;X\; 16)g } X \frac{2 mol}{(5\;mol )} X \frac{26g}{1 mol }$

$m_{_C_2_H_2} = 11.375 g$

Similarly, 5 moles of oxygen are required as a reactant to produce 2 moles of water, thus

$m_{_H_2_O} = 35gX \frac{1 mol}{(2 \;X\; 16)g } X \frac{2 mol}{(5\;mol )} X \frac{18g}{1 mol }$

$m_{_H_2_O} = 7.875 g$

Given that the density of water is $1.0g ml^{-1}$, therefore

$Density = \frac{Mass \;of \;water}{Volume \;of \;water}$

$Volume \;of \;water = \frac{7.875}{1.0 g ml^{-1}}$

The volume of water = 7.875 ml

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Answer 2

Answer:

$a. \ \ \ \text{The mass of ethylene is 11.375 g}. \\ \\ b. \ \ \ \text{The volume of water produced is 7.875 ml}.$

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

           $m_{C_{2}H_{2}} \ = \ 35 \ \text{g} \ \times \ \displaystyle\frac{1 \ \text{mol}}{\left(2 \times 16 \right) \text{g}} \ \times \ \displaystyle\frac{2 \ \text{mol}}{5 \ \text{mol}} \ \times \ \displaystyle\frac{\left(12 \ \times \ 2 \ + \ 1 \ \times \ 2 \right) \ \text{g}}{1 \ \text{mol}} \\ \\ m_{C_{2}H_{2}} \ = \ 11.375 \ \text{g}$

b. Similarly, 5 moles of oxygen are required as a reactant to produce 2  

    moles of water, thus

             $m_{H_{2}O} \ = \ 35 \ \text{g} \ \times \ \displaystyle\frac{1 \ \text{mol}}{\left(2 \times 16 \right) \text{g}} \ \times \ \displaystyle\frac{2 \ \text{mol}}{5 \ \text{mol}} \ \times \ \displaystyle\frac{\left(1 \ \times \ 2 \ + \ 16 \right) \ \text{g}}{1 \ \text{mol}} \\ \\ m_{H_{2}O} \ = \ 7.875 \ \text{g}$

      Given that the density of water is $1.0 \ \text{g ml}^{-1}$, therefore

                                               $\rho_{H_{2}O} \ = \ \displaystyle\frac{m_{H_{2}O}}{V_{H_{2}O}} \\ \\ V_{H_{2}O} \ = \ \displaystyle\frac{m_{H_{2}O}}{\rho_{H_{2}O}} \\ \\ V_{H_{2}O} \ = \ \displaystyle\frac{7.875 \ \text{g}}{1.0 \ \text{g ml}^{-1}} \\ \\ V_{H_{2}O} \ = \ 7.875 \ \text{ml}$