When 50.0 g of iron(3)nitrate, FE(NO3)3, is dissolved in water , the number of nitrate ions present in the solution is
3.73x10^23
1.24 x10^23
To determine the number of nitrate ions present in the solution when 50.0 g of iron(III) nitrate (Fe(NO3)3) is dissolved in water, you need to use the molar mass and stoichiometry of the compound.
Step 1: Find the molar mass of iron(III) nitrate (Fe(NO3)3):
The molar mass of Fe(NO3)3 can be calculated by adding up the atomic masses of its constituent elements. The atomic masses are as follows:
Iron (Fe) = 55.85 g/mol
Nitrogen (N) = 14.01 g/mol
Oxygen (O) = 16.00 g/mol
Molar mass of Fe(NO3)3 = [1 × Fe (55.85 g/mol)] + [3 × N (14.01 g/mol)] + [9 × O (16.00 g/mol)]
= 55.85 + 42.03 + 144.00
= 241.88 g/mol
Step 2: Calculate the number of moles of Fe(NO3)3:
Using the given mass of 50.0 g and the molar mass calculated above, you can determine the number of moles:
Number of moles = Mass of compound / Molar mass
= 50.0 g / 241.88 g/mol
≈ 0.2069 mol
Step 3: Determine the number of nitrate ions:
From the balanced chemical formula of Fe(NO3)3, you can see that there are three nitrate ions (NO3-) for every one iron(III) nitrate (Fe(NO3)3) molecule. Therefore, you need to multiply the number of moles by 3 to get the number of nitrate ions:
Number of nitrate ions = Number of moles of Fe(NO3)3 × 3
≈ 0.2069 mol × 3
≈ 0.6207 mol
So, when 50.0 g of iron(III) nitrate is dissolved in water, the number of nitrate ions present in the solution is approximately 0.6207 moles.