Question

Rust, chemically, is iron(III) oxide,Fe2O3.On a large ship, 16kg of rust accumulates in a day. Calculate how many kilograms of iron and how many kilograms of oxygen would be used up. ( Ans:Fe=4.19 O=.9024 . Give me details)

Answer 1

hint:
the molar mass of \(Fe_2O_3\) is \(159.70\)
So, the number of mol of \(Fe_2O_3\) is:
\[n = \frac{{16000}}{{159.70}} = ...?\]

Answer 2

n=160/159.70=1

Answer 3

hint:
16 Kg are equivalent to 16,000 grams, so we have to do this computation:
\[n = \frac{{16000}}{{159.70}} = ...?\]

Answer 4

n=100.18

Answer 5

correct!

Answer 6

now, 1 mol of \(Fe_2O_3\) contains 2 moles of \(F_e\) and 3 moles of \(O\), so we can write:
\[\Large \begin{gathered}
{m_{Fe}} = \frac{{100.18 \times 2 \times 55.85}}{{1000}} = ...{\text{Kg}} \hfill \\
\hfill \\
{m_O} = \frac{{100.18 \times 3 \times 16}}{{1000}} = ...{\text{Kg}} \hfill \\
\end{gathered} \]

Answer 7

where \(m_O\) and \(m_{Fe}\) are the masses of oxygen and iron respectively

Answer 8

mFe=11.198Kg & mO=4.8

Answer 9

Thanks a lot for solving this

Answer 10

that's right!
As we can see if we compute this sum:
\(11.2+4.8\), we get the initial mass, namely \(16\) Kg

Answer 11

so, the mass is conserved, according to the law of \(Lavoisier\)