A 0.1943g iron ore sample was dissolved in hydrochloric acid and the iron was obtained as Fe 2+ (aq). The iron solution was titrated with Ce 4+ solution according to the balanced chemical reaction shown below.After calculation, it was found that 0.0237 g of iron from the ore reacted with the cerium solution.
Ce 4+(aq)+ Fe2+(aq) --> Ce3+ (aq)+ Fe3+ (aq)
Calculate the mass percent of iron in the original ore sample. Please round your answer to the tenths place
%Fe = (0.0237/mass sample)*100 = ??
12.2
Well, calculating the mass percent of iron in the original ore sample requires a bit of number crunching. Don't worry though, I'll do my best to keep it entertaining!
First, let's start by finding the number of moles of iron in 0.0237 g of iron. Since the molar mass of iron is 55.85 g/mol, we can use the formula:
moles of iron = mass of iron / molar mass of iron
moles of iron = 0.0237 g / 55.85 g/mol
Now, let's use the balanced chemical equation to determine the moles of cerium required to react with the moles of iron. According to the equation, it's a 1:1 ratio.
moles of cerium = moles of iron
Now that we have the moles of cerium, we can determine the mass of cerium using its molar mass, which is 140.12 g/mol.
mass of cerium = moles of cerium x molar mass of cerium
mass of cerium = 0.0237 mol x 140.12 g/mol
Finally, we can calculate the mass percent of iron in the original ore sample using the following formula:
mass percent of iron = (mass of iron / mass of ore sample) x 100%
Remember, the mass of the ore sample was given as 0.1943 g.
mass percent of iron = (0.0237 g / 0.1943 g) x 100%
Now, that should give you the mass percent of iron in the original ore sample. Just remember to round your answer to the tenths place!
I hope my calculations have at least brought a smile to your face!
To calculate the mass percent of iron in the original ore sample, we need to determine the amount of iron present in the 0.1943 g iron ore sample, and then calculate the mass percent using the given information.
First, let's find the molar ratio between Ce4+ and Fe2+ from the balanced chemical reaction:
Ce4+(aq) + Fe2+(aq) → Ce3+(aq) + Fe3+(aq)
From the balanced equation, we can see that 1 mole of Ce4+ reacts with 1 mole of Fe2+. This means that the moles of Ce4+ solution used in the titration reaction is equal to the moles of Fe2+ reacted.
To determine the moles of iron reacted, we need to convert the given mass of iron reacted (0.0237 g) to moles. We do this by dividing the mass of iron by its molar mass.
The molar mass of iron (Fe) is 55.845 g/mol. So, the moles of iron reacted are:
moles of Fe = mass of Fe reacted / molar mass of Fe
moles of Fe = 0.0237 g / 55.845 g/mol
Next, we can calculate the mass percent of iron in the original ore sample using the following formula:
mass percent of iron = (mass of Fe / mass of ore sample) * 100
Since we have already calculated the moles of Fe, we can find the mass of Fe:
mass of Fe = moles of Fe * molar mass of Fe
mass of Fe = (0.0237 g / 55.845 g/mol) * 55.845 g/mol
Finally, we can calculate the mass percent of iron in the original ore sample:
mass percent of iron = (mass of Fe / mass of ore sample) * 100
mass percent of iron = [(0.0237 g / 55.845 g/mol) * 55.845 g/mol] / 0.1943 g * 100
Simplifying the expression:
mass percent of iron = (0.0237 g / 0.1943 g) * 100
Calculating the value:
mass percent of iron = 12.2% (rounding to the tenths place)
Therefore, the mass percent of iron in the original ore sample is approximately 12.2%