A 46.9 g of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 11.6 grams. What was the percent iron in the sample of ore?
Answer in units of %
Convert 11.6 g Fe2O3 to g Fe. The easy way to do that is to multiply by a chemical factor which, which as far as I know, isn't taught anywhere in the world anymore anymore.
11.6 x (2*molar mass Fe/molar mass Fe2O3) = g Fe.
Then %Fe = (g Fe/mass ore sample)*100 = ??
To find the percent iron in the sample of ore, you need to calculate the ratio of the mass of iron to the mass of the ore and then convert it to a percentage.
First, calculate the mass of iron in the sample by subtracting the mass of Fe2O3 from the mass of the ore:
Mass of iron = Mass of ore - Mass of Fe2O3
Mass of iron = 46.9 g - 11.6 g
Mass of iron = 35.3 g
Next, calculate the percent iron by dividing the mass of iron by the mass of the ore and then multiplying by 100:
Percent iron = (Mass of iron / Mass of ore) x 100
Percent iron = (35.3 g / 46.9 g) x 100
Percent iron = 0.7529 x 100
Percent iron = 75.29%
Therefore, the percent iron in the sample of ore is 75.29%.
To calculate the percent of iron in the sample of ore, you can use the formula:
Percent (%) iron = (mass of Fe / mass of ore) x 100
Given data:
Mass of iron (Fe) = 11.6 grams
Mass of ore = 46.9 grams
Let's substitute the values into the formula and solve:
Percent iron = (11.6 g / 46.9 g) x 100
Percent iron = 0.247 grams / 46.9 grams x 100
Percent iron = 24.7%
Therefore, the percent of iron in the sample of ore is 24.7%.