If 13.75 g of iron (III) nitrate react with an excess of ammonium sulfide, how many grams of iron (III) sulfide are produced? Report your answer to 2 sig figs

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To determine the number of grams of iron (III) sulfide produced, you need to first determine the balanced chemical equation for the reaction between iron (III) nitrate and ammonium sulfide:

Fe(NO3)3 + (NH4)2S → Fe2S3 + 3NH4NO3

From the balanced equation, you can see that one mole of iron (III) nitrate reacts to produce one mole of iron (III) sulfide. To convert grams of iron (III) nitrate to moles, you need to use its molar mass.

The molar mass of iron (III) nitrate (Fe(NO3)3) can be calculated by adding up the molar masses of its constituent elements:
(1 x molar mass of Fe) + (3 x molar mass of N) + (9 x molar mass of O)

Fe: 55.845 g/mol
N: 14.007 g/mol
O: 16.00 g/mol

Molar mass of Fe(NO3)3 = (1 x 55.845) + (3 x 14.007) + (9 x 16.00) g/mol

Now that you have the molar mass of iron (III) nitrate, you can calculate the number of moles using the given mass of 13.75 g and the following formula:

moles = mass / molar mass

Then, using the balanced equation, you can determine the moles of iron (III) sulfide produced, which is the same as the moles of iron (III) nitrate.

Finally, you can convert the moles of iron (III) sulfide back into grams by multiplying by its molar mass, which can be calculated the same way as for iron (III) nitrate.

Remember to report your answer to 2 significant figures.