Answer
There are 49.2 g of chromium in 234 g of that ore.
75.7 g of pure chromium is obtained from the recovery process.
Work Step by Step
Chromium mass in 1 mol of $FeCr_2O_7$:
$mm(Cr)* 2 = 52.00 *2 = 104.0g(Cr)/mol(FeCr_2O_7)$
Molar mass $(FeCr_2O_7)$: $55.85* 1 + 52.00* 2 + 16.00* 7 = 271.85g/mol$
Purity of the ore: 55.0% $FeCr_2O_7$ by mass.
- Use these informations as conversion factors:
$234$ $g (ore) \times \frac{55.0\%(FeCr_2O_7)}{100\%(ore)} \times \frac{1 mol(FeCr_2O_7)}{271.85g (FeCr_2O_7)} \times \frac{104.0g(Cr)}{1mol(FeCr_2O_7)} = 49.2g (Cr)$
------------------
Find the mass of chromium in 400 g of the ore:
$400$ $g (ore) \times \frac{55.0\%(FeCr_2O_7)}{100\%(ore)} \times \frac{1 mol(FeCr_2O_7)}{271.85g (FeCr_2O_7)} \times \frac{104.0g(Cr)}{1mol(FeCr_2O_7)} = 84.16g (Cr)$
- Now, use the 90.0% factor to find the chromium that can be recovered.
$84.16g(Cr) \times \frac{90.0\% (Recovered-Cr)}{100\%(Cr)} = 75.7g (Recovered-Cr)$
