Chem Help Please. I asked this question the other day but still no help. I appreciate any help please.?

Question

1. Iron metal can be produced from the mineral hematite, Fe2O3, by reaction with carbon:

Fe2O3(s) + C(s) --> Fe(s) + CO(g)
(a) Balance the equation.
(b) How may moles of carbon are needed to reach with 525 g of hematite?
(c) How may grams of carbon ar eneede dto react with 525 g of hematite?

2. In the preparation of iron from hematite (above question), how many kilograms of iron are present in 105 kg of hematite?

3. We saw in (question 1) that iron metal can be made by heating iron ore with carbon according to the (unbalanced) equation:
Fe2O3(s) + C(s) --> Fe(s) + Co2(g)
(a) How many kilograms of iron can be obtained from the reaction of 75.0 kg of Fe2O3?
(b) If 51.3 kg of iron is obtained from the reaction in part (a), what is the percent yield?

4. Which of the following substances are likely to be soluble in water?
(a) ZnS
(b) Au2(Co3)3
(c) PbCl2
(d) MnO2

Answer 1

1a) 2Fe(s) in products because atoms in must be the same as atoms out. Enough CO in products to account for all the O in Fe2O3. Then enough C in reactants to account for all the C in the CO product.

1b) you always need to convert to moles, when going from one substance to another.

Divide 525g by the formula weight of hematite to get the number of laws of hematite. Then multiply that by the number of C that react with each hematite (you found that out in the last part)

1c) You now know moles of carbon. Multiply by molar mass of carbon to get grounds of carbon.

2) 105 kg hematite divided by molar mass of hematite is number of kilomoles (I just made that word up, but you get the idea) of hematite. Each kilomole of hematite contains 2 kilomoles iron. Multiply by molar mass of iron to get kg iron.

3) Now you've got me confused! The processes described by the equation in question 1 and by the equation in this question both happen, but which one do you mean?

However, the calculation here he is exactly the same as a calculation in (2).

Percentage yield equals actual yield divided by calculated (theoretical) yield.

4 This is something you will find in your chemistry textbook.