## b) To calculate the theoretical mass of copper that was expected to be produced, we need to use stoichiometry, which is the study of relative quantities of reactants and products in chemical reactions.

First, we need to determine the balanced chemical equation for the reaction:

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

From the balanced chemical equation, we can see that the molar ratio between Fe and Cu is 1:1. This means that for every 1 mole of Fe reacted, 1 mole of Cu should be produced.

To calculate the theoretical mass of Cu, we need to know the molar mass of Cu. The molar mass of Cu is 63.55 g/mol.

Since the balanced equation tells us that the molar ratio of Fe to Cu is 1:1, we can directly convert the mass of Fe reacted to the theoretical mass of Cu produced.

Calculations:

1. Convert the mass of Fe to moles:

Moles of Fe = mass of Fe / molar mass of Fe = 0.9845 g / 55.85 g/mol = 0.0176 mol

2. Convert moles of Fe to moles of Cu:

Moles of Cu = moles of Fe (according to the balanced equation) = 0.0176 mol

3. Convert moles of Cu to mass of Cu:

Mass of Cu = moles of Cu x molar mass of Cu = 0.0176 mol x 63.55 g/mol = 1.118 g

Therefore, the theoretical mass of copper expected to be produced is 1.118 g.

Now, to determine the percentage yield of Joel's experiment, we need to compare the actual mass of copper obtained to the theoretical mass.

Calculations:

Actual mass of Cu obtained = 1.024 g (given)

Percentage yield = (Actual mass / Theoretical mass) x 100

= (1.024 g / 1.118 g) x 100

= 91.6%

Therefore, the percentage yield of Joel's experiment is 91.6%.

c) To calculate the number of molecules of iron in 0.9845 g of iron fillings, we need to use the concept of molar mass and Avogadro's number.

1. Calculate the molar mass of iron (Fe):

Molar mass of Fe = 55.85 g/mol

2. Convert the mass of Fe to moles:

Moles of Fe = mass of Fe / molar mass of Fe

= 0.9845 g / 55.85 g/mol

= 0.0176 mol

3. Convert moles to molecules using Avogadro's number:

Number of molecules = Moles of Fe x Avogadro's number

= 0.0176 mol x 6.022 x 10^23 molecules/mol

= 1.06 x 10^22 molecules

Therefore, there are approximately 1.06 x 10^22 molecules of iron in 0.9845 g of iron fillings.