Final answer:
To calculate the grams of Fe(OH)3 from 3.6 moles of NaOH, we use a balanced equation, molar ratios, and the molar mass of Fe(OH)3. Following the stoichiometry, 3.6 moles of NaOH yield 1.2 moles of Fe(OH)3, which equates to approximately 128 grams of Fe(OH)3 when using its molar mass of 106.87 g/mol.
Step-by-step explanation:
The student asked how many grams of Fe(OH)3 form from 3.6 moles of NaOH. To solve this problem, we need the balanced chemical equation that relates Fe(OH)3 formation with NaOH.
However, based on the reaction provided, which is 3NaOH(aq) + Fe(NO3)3(aq) → Fe(OH)3(s) + 3NaNO3(aq), we can observe that 3 moles of NaOH are required to produce 1 mole of Fe(OH)3.
This means that in our problem, 3.6 moles of NaOH would produce 3.6/3 = 1.2 moles of Fe(OH)3. The molar mass of Fe(OH)3 is calculated by adding the molar masses of Fe, O, and H accordingly:
1x55.845 g.mol−¹ Fe + 3x15.999 g.mol−¹ O + 3x1.008 g.mol−¹ H, equals to roughly 106.87 g.mol−¹ of Fe(OH)3. Therefore, the mass of Fe(OH)3 produced is 1.2 moles * 106.87 g/mol,
which equals to approximately 128.24 grams of Fe(OH)3. This value should be reported to three significant figures, 128 grams.