If 13.75 g of Iron (III) nitrate react with an excess of ammonium sulfide, how many grams of iron(III) sulfide are produced?

To determine the number of grams of iron(III) sulfide produced, we first need to write and balance the chemical equation for the reaction between iron(III) nitrate and ammonium sulfide.

The chemical equation is:

Fe(NO3)3 + (NH4)2S → Fe2S3 + 6NH4NO3

From the balanced equation, we can deduce that the stoichiometric ratio of Fe(NO3)3 to Fe2S3 is 1:1. This means that 1 mole of Fe(NO3)3 reacts to produce 1 mole of Fe2S3.

To find the number of moles of Fe2S3 produced, we need to convert the given mass of Fe(NO3)3 into moles.

The molar mass of Fe(NO3)3 is:
Fe: 55.85 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3 because of three oxygen atoms)
Total molar mass of Fe(NO3)3 = 55.85 + (14.01 * 3) + (16.00 * 3) = 241.99 g/mol

To find the number of moles of Fe(NO3)3, we divide the given mass by the molar mass:
moles of Fe(NO3)3 = 13.75 g / 241.99 g/mol ≈ 0.05678 mol

Since the stoichiometric ratio is 1:1, the number of moles of Fe2S3 produced is also approximately 0.05678 mol.

Now, to determine the mass of Fe2S3, we need to multiply the number of moles by its molar mass.

The molar mass of Fe2S3 is:
Fe: 55.85 g/mol (x2 because of two iron atoms)
S: 32.07 g/mol (x3 because of three sulfur atoms)
Total molar mass of Fe2S3 = (55.85 * 2) + (32.07 * 3) = 207.81 g/mol

Mass of Fe2S3 = moles of Fe2S3 * molar mass of Fe2S3
Mass of Fe2S3 = 0.05678 mol * 207.81 g/mol ≈ 11.82 g

Therefore, approximately 11.82 grams of iron(III) sulfide are produced.

1. Write and balance the equation

2. Convert 3.75 g iron(III) nitrate to mols. mols = grams/molar mass
3. Using the coefficients in the balanced equation, convert mols iron(III) nitrate to mols of the product.
4. Finally, convert mols of the product to grams. g = mols x molar mass.
This 4-step procedure will work all simple (as opposed to limiting reagent problems) stoichiometry problems.