If 10 g of C2H2 is mixed with 80 g of Br2, what masses of c2h2br2 and c2h2br4 will be formed? Assume that all the C2H2 has reacted.

If 10 g of C2H2 is mixed with 80 g of Br2, what masses of c2h2br2 and c2h2br4 will be formed? Assume that all the C2H2 has reacted.

To determine the masses of C2H2Br2 and C2H2Br4 formed, we need to understand the chemical reaction between C2H2 and Br2.

The reaction is as follows:

C2H2 + Br2 → C2H2Br2 (Dibromoethene)
C2H2Br2 + Br2 → C2H2Br4 (Tetrabromoethane)

First, calculate the number of moles of C2H2 and Br2 present in the given amounts.

Molar mass of C2H2 (Ethyne) = 2(12.01 g/mol) + 2(1.01 g/mol) = 26.04 g/mol
Molar mass of Br2 = 2(79.90 g/mol) = 159.80 g/mol

Number of moles of C2H2 = mass / molar mass = 10 g / 26.04 g/mol ≈ 0.384 mol
Number of moles of Br2 = mass / molar mass = 80 g / 159.80 g/mol ≈ 0.501 mol

Now, let's calculate the amount of C2H2Br2 formed:

From the balanced chemical equation, we can see that the stoichiometric ratio between C2H2 and C2H2Br2 is 1:1.

Therefore, the number of moles of C2H2Br2 formed will be equal to the number of moles of C2H2.

Number of moles of C2H2Br2 = 0.384 mol

Now, let's calculate the amount of C2H2Br4 formed:

From the balanced chemical equation, we can see that the stoichiometric ratio between C2H2Br2 and C2H2Br4 is 1:1.

Therefore, the number of moles of C2H2Br4 formed will be equal to the number of moles of C2H2Br2, which is 0.384 mol.

Finally, calculate the masses of C2H2Br2 and C2H2Br4 formed using their respective molar masses.

Mass of C2H2Br2 = number of moles × molar mass = 0.384 mol × (1(12.01 g/mol) + 2(79.90 g/mol)) ≈ 37.58 g
Mass of C2H2Br4 = number of moles × molar mass = 0.384 mol × (1(12.01 g/mol) + 4(79.90 g/mol)) ≈ 182.21 g

Therefore, approximately 37.58 g of C2H2Br2 and 182.21 g of C2H2Br4 will be formed when 10 g of C2H2 is mixed with 80 g of Br2.