13.75 g of iron (III) nitrate react with an excess of ammonium sulfide, how many grams of iron (III) sulfide are produced
Done the same way as the CO2 problem above.
To determine the number of grams of iron (III) sulfide produced, we need to calculate the molar ratio between iron (III) nitrate and iron (III) sulfide and then use stoichiometry.
Step 1: Write the balanced equation for the reaction.
Fe(NO3)3 + (NH4)2S -> Fe2S3 + 6NH4NO3
Step 2: Find the molar mass of iron (III) nitrate.
Fe(NO3)3: Fe = 55.845 g/mol | N = 14.007 g/mol (3 atoms) | O = 15.999 g/mol (9 atoms)
Total molar mass of Fe(NO3)3 = 55.845 + (14.007 * 3) + (15.999 * 9) = 241.858 g/mol
Step 3: Calculate the number of moles of iron (III) nitrate.
moles of Fe(NO3)3 = mass of Fe(NO3)3 / molar mass of Fe(NO3)3
moles of Fe(NO3)3 = 13.75 g / 241.858 g/mol = 0.057 moles
Step 4: Determine the molar ratio between iron (III) nitrate and iron (III) sulfide.
From the balanced equation, the ratio of Fe(NO3)3 to Fe2S3 is 1:1.
Step 5: Calculate the mass of iron (III) sulfide.
mass of Fe2S3 = moles of Fe(NO3)3 * molar mass of Fe2S3
molar mass of Fe2S3 = (55.845 * 2) + (32.06 * 3) = 207.9 g/mol
mass of Fe2S3 = 0.057 moles * 207.9 g/mol = 11.8623 g
Therefore, approximately 11.86 grams of iron (III) sulfide are produced in the reaction.