Given the following dissolving process: Fe2(SO4)3 (s) = 2Fe3+ (aq) + 3SO42- (aq) How many grams of iron (III) ions are produced when 0.450 moles of sulfate ion are produced?
options:
50.3 g
16.8 g
50.3 g
37.7 g
75.4 g
same stoichiometry process.
so whts th answer from th options??
To solve this problem, we need to use the stoichiometry (molar ratios) from the balanced chemical equation to convert moles of sulfate ions to moles of iron (III) ions, and then convert moles to grams using the molar mass.
From the balanced chemical equation:
1 mole of Fe2(SO4)3 produces 2 moles of Fe3+ ions
Therefore, 1 mole of Fe3+ ions is produced from (1/2) moles of Fe2(SO4)3.
Given:
0.450 moles of sulfate ion (SO42-), which is the same as 0.450 moles of Fe2(SO4)3 since the stoichiometric coefficient for SO42- is 1.
To calculate the moles of Fe3+ ions:
0.450 moles of Fe2(SO4)3 × (1/2) moles of Fe3+ ions / 1 mole of Fe2(SO4)3 = 0.225 moles of Fe3+ ions
Next, we need to convert moles of Fe3+ ions to grams using the molar mass of Fe3+ ions.
The molar mass of Fe3+ ions can be calculated as follows:
Molar mass of Fe3+ ions = molar mass of Fe × 3
= 55.845 g/mol × 3
= 167.535 g/mol
To calculate the grams of Fe3+ ions:
0.225 moles of Fe3+ ions × 167.535 g/mol = 37.7 g
Therefore, the answer is 37.7 g.
The correct option is:
- 37.7 g