Question

Searches related to If 0.75 grams of iron (Fe) react according to the following reaction, how many grams of copper (Cu) will be produced? Fe + CuSO4 -> Cu + FeSO4

Answer 1

Answer:

0.83 g

Explanation:

Step 1: Write the balanced equation

Fe + CuSO₄ ⇒ Cu + FeSO₄

Step 2: Calculate the moles corresponding to 0.75 g of Fe

The molar mass of Fe is 55.85 g/mol.

$0.75g \times \frac{1mol}{55.85g} = 0.013 mol$

Step 3: Calculate the moles of Cu produced from 0.013 moles of Fe

The molar ratio of Fe to Cu is 1:1. The moles of Cu produced are 1/1 × 0.013 mol = 0.013 mol.

Step 4: Calculate the mass corresponding to 0.013 moles of Cu

The molar mass of Cu is 63.55 g/mol.

$0.013mol \times \frac{63.55g}{mol} = 0.83 g$

Answer 2

Answer:

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.

Explanation:

You know the following balanced reaction:

Fe + CuSO₄ ⇒ Cu + FeSO₄

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities react and are produced:

• Fe: 1 mole
• CuSO₄: 1 mole
• Cu: 1 mole
• FeSO₄: 1 mole

Being:

• Fe: 55.85 g/mole
• Cu: 63.54 g/mole
• S: 32 g/mole
• O: 16 g/mole

the molar mass of the compounds participating in the reaction is:

• Fe: 55.85 g/mole
• CuSO₄: 63.54 g/mole + 32 g/mole+ 4* 16 g/mole= 159.54 g/mole
• Cu: 63.54 g/mole
• FeSO₄: 55.85 g/mole + 32 g/mole+ 4* 16 g/mole= 151.85 g/mole

Then, by stoichiometry of the reaction, the amounts of reagent and product that participate in the reaction are:

• Fe: 1 mole*55.85 g/mole= 55.85 g
• CuSO₄: 1 mole* 159.54 g/mole= 159.54 g
• Cu: 1mole* 63.54 g/mole= 63.54 g
• FeSO₄: 1 mole* 151.85 g/mole= 151.85 g

Then you can apply a rule of three as follows: if 55.85 grams of Fe produces 63.54 grams of Cu, 0.75 grams of Fe how much mass of Cu does it produce?

$mass of Cu=\frac{0.75 grams of Fe*63.54 grams of Cu}{55.85 grams of Fe}$

mass of Cu= 0.85 grams

If 0.75 grams of iron (Fe) react, 0.85 grams of copper (Cu) will be produced.