Question

QUESTION 1

If 12.7 moles of Cu and 15.7 moles of HNO3 are allowed to react, what is the maximum number of moles of NO that can be produced?

3 Cu + 8 HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

QUESTION 2

What mass of aluminum oxide can be recovered from the complete reaction of 31.1 g of Al with 31.1 g of oxygen gas?

4 Al (s) + 3 O2(g) → 2 Al2O3(s)

QUESTION 3

If 8.7 moles of Al and 2.7 moles of Fe2O3 are allowed to react completely, how many moles of the excess reactant will remain? (Please give your answer with one decimal place even if one of the initial values does not show any decimal places.)

2Al + Fe2O3 → Al2O3 + 2Fe

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Answer 1

1)

3 Cu   +       8 HNO3 -------------------> 3Cu(NO3)2 + 2NO + 4H2O

3 mol           8                                                         2

12.7 mol        15.7                                                     ??

here limiting reagent is HNO3. because the mol ratio of HNO3 is less.

8 mol HNO3   -----------------> 2 mol NO

15.7 mol HNO3    -------------->   ??

moles of NO = 15.7 x 2 / 8 = 3.92 mol

moles of NO = 3.92 mol

2)

4 Al (s) + 3 O2(g) → 2 Al2O3(s)

108 g       96 g              203.92 g

31.1 g      31.1 g              ??

here limiting reagent is Al .

108 g Al    ---------------> 203.92 g Al2O3

31.1 g Al    ---------------->   ??

mass of Al2O3 = 31.1 x 203.92 / 108 = 58.7 g

mass of Al2O3 = 58.7 g