## Chemistry 10th Edition

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. $84.16g(Cr) \times \frac{90.0\% (Recovered-Cr)}{100\%(Cr)} = 75.7g (Recovered-Cr)$