A 0.1300 g iron ore sample was dissolved in hydrochloric acid and the iron was obtained as Fe2+(aq). The iron solution was titrated with Ce4+ solution according to the balanced chemical reaction shown below. After calculation, it was found that 0.0259 g of iron from the ore reacted with the cerium solution.

Ce4+(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

0.0259 x 100/ 0.1300 = 19.9%

I think i did this problem correctly, but I wanted it to be approved by an expert.

I agree.

It seems like you made a calculation mistake in your answer. Let's go through the correct steps to calculate the mass percent of iron in the original ore sample.

First, we need to determine the number of moles of Fe that reacted with the Ce4+ solution. To do this, we can use the molar mass of Fe. The molar mass of Fe is 55.845 g/mol.

Number of moles of Fe = mass of Fe / molar mass of Fe = 0.0259 g / 55.845 g/mol = 0.000463 mol

Next, we can calculate the number of moles of Fe in the original ore sample. Since the balanced chemical reaction tells us that the ratio of Fe2+ to Fe in the reaction is 1:1, the number of moles of Fe in the original ore sample is also 0.000463 mol.

Now, we need to calculate the mass of the original ore sample. We already know that the mass of the Fe in the ore sample is 0.0259 g. Since the ratio of Fe to the ore sample is 0.000463 mol : 0.000463 mol = 1:1, the mass of the original ore sample is also 0.0259 g.

Finally, we can calculate the mass percent of iron in the original ore sample.

Mass percent of iron = (mass of Fe / mass of original ore sample) x 100
Mass percent of iron = (0.0259 g / 0.0259 g) x 100 = 100.0%

Therefore, the mass percent of iron in the original ore sample is 100.0%.

Note: It is always a good practice to double-check your calculations and make sure you have considered all the given information and stoichiometry of the reaction.