The rusting of iron is represented by the equation 4Fe +3O2 --> 2Fe2O3. If you have a 1.50-mol sample of iron, how many moles of Fe2O3 will there be after the iron has rusted completely?

Look at the equation.

The equation tells us that 4 moles Fe will produce 1 moles of the oxide; therefore, 1.5 mol will produce 1/2 that or 0.75 mol which is answer b.

Thank you very much!

I tried answering this one but I am getting answers that are not in the 4 choices given...Help?

8. The balanced equation P4 (s) + 6H2 (g) --> 4PH3 tells us that 2 mol H2

a. reacts with 1 mol 4
b. produces 4 mol PH3
c. cannot react with phosphorus
d. produces 2 mol PH3
e. reacts with 2 mol P4

Thank you!!!

Well, let's rustle up an answer for you! The balanced equation you provided shows that 4 moles of iron (Fe) react to form 2 moles of iron(III) oxide (Fe2O3). So, for every 4 moles of Fe, we get 2 moles of Fe2O3.

Since you started with a 1.50-mol sample of iron, we need to figure out how many moles of Fe2O3 that would form. Using the ratio from the equation, for every 4 moles of Fe, we would expect 2 moles of Fe2O3.

Therefore, we can calculate the answer by dividing 1.50 moles of Fe by 4 and then multiplying by 2: (1.50 mol Fe / 4) * 2 = 0.75 mol Fe2O3.

So your initial deduction of "b" is spot on! The correct answer is indeed 0.75 mol of Fe2O3. Great work!

Yes, to determine the correct answer, we can use the stoichiometry of the balanced chemical equation. The equation shows that 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.

Since we have 1.50 moles of Fe, we can set up a proportion to calculate the moles of Fe2O3:

(1.50 mol Fe) x (2 mol Fe2O3 / 4 mol Fe) = 0.75 mol Fe2O3

So, you are correct. The correct answer is option b) 0.75 mol.