The reaction can be described using the equation: 2C2H25O24CO22H2O. How much C2H2is needed to react with 68.1 g of O2to produce 75.0g of CO2and 15.3 grams of steam?

22.13g

Explanation:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

2C2H2 + 5O2 —> 4CO2 + 2H2O

Next, we'll calculate the mass of C2H2 and O2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of C2H2 = (12x2) + (2x1)

= 24 + 2 = 26g/mol

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

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

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

Now, we can obtain the mass of C2H2 that will react with 68.1g of O2 as follow:

From the balanced equation above,

52g of C2H2 reacted with 160g of O2.

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

Xg of C2H2 = (52x68.1)/160

Xg of C2H2 = 22.13g

Therefore, 22.13g of C2H2 is needed to react with 68.1g of O2

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%