According to the reaction Cl2O7 + H2O → 2HClO4, how many grams of Cl2O7 must react with an excess of H2O to form 56.2 g HClO4?

51.17g of Cl2O7

Explanation:

The equation for the reaction is given below:

Cl2O7 + H2O → 2HClO4

First, let us calculate the masses of Cl2O7 and HClO4 from the balanced equation. This is illustrated below:

Molar Mass of Cl2O7 = (2x35.5) + (16x7) = 71 + 112 = 183g/mol

Molar Mass of HClO4 = 1 + 35.5 + (16x4) = 1 + 35.5 + 64 = 100.5g/mol

Mass of HClO4 from the balanced equation = 2 x 100.5 = 201g

Now, we can obtain the mass of Cl2O7 that must react to form 56.2g HClO4. This is illustrated below:

From the equation above,

183g of Cl2O7 reacted to produced 201g of HClO4.

Therefore, Xg of Cl2O7 will react to produce 56.2g of HClO4 i.e

Xg of Cl2O7 = (183 x 56.2)/201

Xg of Cl2O7 = 51.17g

From the calculations made above, 51.17g of Cl2O7 is required to produce 56.2g of HClO4

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.

your answer is c) plasma

~*"auba14"*~