Question #b5376

Question

2584 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-15T04:41:01

$8.3997$ x $10^25$ atoms of $Cu$

We need to balance the equation to determine the mole ratio of $Cu$ against the $Cu_3FeS_3$ ;
The balanced equation is:
$2Cu_3FeS_3 + 7O_2 -> 6Cu = 2FeO + 6SO_2$
Find the formula mass of bornite, by locating the atomic masses of the elements composing the copper ore, which is equal to 344 grams per mole;
To start the calculation, you are provided with the mass of the raw material of $20 kg$ which has 80% purity. Meaning $80%$ of the raw material is bornite (copper ore). To find the mass of copper ore, multiply the mass by its purity; so that
$20 kg$ x $0.80=16 kg$ of pure bornite
Convert the mass (kg) of pure bornite to grams to be consistent with the units being used in the formula mass; ( $16,000$ grams)
Again, $16,000$ grams bornite will be converted to mole bornite by multiplying with the mole ratio reflected on the balanced equation above to arrive $139.53$ mole $Cu$ ;
Since $1 mol Cu=6.02$ x $10^23$ atoms $Cu$ , multiply the answer in step 6 by the conversion factor taken from this equaivalent statement, which is $(6.02 x 10^23 atms Cu)/(1 mol Cu)$ ;
The answer is computed as 8.3997 x $10^25$ atoms $Cu$