An iron ore sample is impure Fe2O3. When Fe2O3 is heated with excess carbon, iron metal is produced. A 7.720 g sample of ore yields 4.53 g of Fe. What is the percent, by mass, of Fe2O3 in the ore?
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Your question is somewhat complicated. What are those impurities ('different ingredients')? How most of the impurity is interior the iron ore pattern? as a fashion to calculate the parts you seek for you may desire to comprehend those issues. If we use 0.a hundred and fifty kg we are able to basically obtain the utmost achieveable grams of fe and CO2 that could desire to be obtained from a organic pattern.
Figure out how many moles of Fe you got. 4.53 g (1 mole/55.8g) 0.0812 moles Fe Look at the reaction: Fe2O3 + C --- 2Fe + CO2, then balance: 2Fe2O3 + 3C --- 4Fe + 3CO2 You can see from the balanced equation that you get 2 moles of Fe from every 1 mole of Fe2CO3 So you needed 1/2 (0.0812 moles) 0.0406 moles of Fe2CO3 Find the molar mass of Fe2CO3 2(55.8) + 12.0 + 3 (16.0) 111.6 +12.0 +48.0 171.6 Find the actual mass of Fe2CO3. 0.0406 moles(171.6 g/mole0 6.97 g Fe2CO3 % in the ore (6.97 g/7.72 g) x 100% 90.2%
