Question

Assume that it was necessary to produce 8 g of copper. How much iron would you have to use?

Answer 1

First, we have to check if our mass balance is correct:

We calculate the mass of obtained copper:

`m\text{ }_(Cu)=\text{ 2.4581 g - 0.6503 g=1.8078 g}`

Now, we transform our data into moles, with the respective molecular weights:

`\begin{gathered} M.W\text{ of CuSO}_4=\text{ 64+32+16*4=160 g/mol} \\ M.W.\text{ of iron = 56 g/mol} \\ M.W.\text{ of cupper: 64 g/mol} \end{gathered}`
`\begin{gathered} 12.5056\text{ g CuSO}_4*\text{ }\frac{1\text{ mole}}{160\text{ g}}=0.07816\text{ moles CuSO}_4 \\ \\ 1.4976\text{ g steel *}\frac{98\text{ g Fe}}{100\text{ g steel}}*\frac{1\text{ mole Fe}}{56\text{ g Fe}}=0.02621\text{ moles Fe} \end{gathered}`

As the relation is 1:1, we can say that the limiting reactant is Fe. We calculate the theoretical mass of produced copper, based on the moles of Fe.

`0.02621\text{ moles Fe *}\frac{1\text{ mole Cu}}{1\text{ mole Fe}}*\frac{64\text{ g Cu}}{1\text{ mole Cu}}=\text{ 1.6774 g Cu}`

Now, we can say that the experiment is well performed as the theoretical mass of cupper (1.6774 g) is very similar to the obtained one (1.8078 g).

And now, we will calculate the requirement of the exercise using this theoretical form:

`8\text{ g Cu * }\frac{1\text{ mole Cu}}{64\text{ g Cu}}*\frac{1\text{ mole Fe}}{1\text{ mole Cu}}*\frac{56\text{ g Fe}}{1\text{ mole Fe}}=\text{ 7 g Fe}`

And we could also calculate the quantity of steel if we assume the concentration of iron at 98% again:

`7\text{ g Fe * }\frac{100\text{ g steel}}{98\text{ g Fe}}=7.1429\text{ g steel}`

The answer is that we need 7 g of iron or 7.1429 g of steel.