Acetylene gas (C2H2) is used in welding torches. When it reacts with oxygen, it produces carbon dioxide (CO2) and steam (H2O). The reaction can be described by the equation:
2C2H2 + 5O2 ➔ 4CO2 + 2H2O. How much mass of C2H2 is needed to react with 68.1 g of O2 to produce 75.0 g of CO2 and 15.35 g of steam?

Explanation:

First thing to understand is how you can manipulate the Ideal Gas Law to your needs. since at a constant temperature, constant. That means,

If we say that subscript 1= oxygen and subscript 2= acetylene, then for acetylene,

Next, we realize from the reaction equation given above that 2 moles of acetylene (n 2 ) react with 5 moles of oxygen (n1). We can plug that right into our equation for P 2, along with the rest of the conditions given in the question:

22.13g of C2H2

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2C2H2 + 5O2 ➔ 4CO2 + 2H2O

Step 2:

Determination of masses of C2H2 and O2 that reacted from the balanced equation. We need to obtain the mass of C2H2 and that of O2 that reacted based on the chemical equation, as it will enhance the process of obtaining the desired result. This is illustrated below:

2C2H2 + 5O2 ➔ 4CO2 + 2H2O

Molar Mass of C2H2 = (12x2) + (2x1) = 24 + 2 = 26g/mol

Mass of C2H2 from the balanced equation = 2 x 26 = 52g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g

From the balanced equation,

52g of C2H2 reacted with 160g of O2.

Step 3:

Determination of the mass of C2H2 needed to react with 68.1g of O2.

From the balanced equation,

52g of C2H2 reacted with 160g of O2

Therefore, Xg of C2H2 will react with 68.1g of O2 i.e

Xg of C2H2 = (52 x 68.1)/160

Xg of C2H2 = 22.13g

From the calculations made above, 22.13g of C2H2 is needed to react with 68.1 g of O2 to produce 75.0 g of CO2 and 15.35 g of steam

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