Answer to Question #86093 in Organic Chemistry for elion

Question

In the CuSO4 solution, 20 g of Fe was thrown, as the solution was dried and weighed 22g. How did the Fe react in the solution?

Answer 1

Iron is more active metal than copper, so iron displaces copper in its salts. In the described task iron dissolves while copper deposits on the surface of the iron. The equation for this reaction:

If we need to know, how much of the iron dissolved and how much of the copper deposited on the surface, we need to make the equation. Let the x be the number of dissolved moles of iron. The mass of such amount of iron in grams will be 56x (as the molar mass of iron is 56 g/mol). At the same time, the same amount of copper (because of the quotients before Fe and Cu are equal to 1) deposits on iron. Mathematically mass of the copper will look like 64x. And as the iron dissolves and reduces its mass, we need to place minus sign next to 56x. The final equation looks like:

64x-56x=2; (2 grams is the difference between starting and ending mass)

8x=2;

x=0.25 moles;

This means that during reaction 0.25 moles (or 56 g/mol * 0.25=14 grams) of iron dissolved and 0.25 moles of copper (or 64 g/mol * 0.25 mol = 16 grams) deposited on the surface of iron.

Answer:

During mentioned process 14 grams of iron dissolves and 16 grams of copper deposits on the surface of the iron.