chemistry help!?

Question

What mass of SO3 is produced from the reaction b/t 31.5g S8 and 8.65g of O2?

What mass of H2SO4 is produced from the reaction of 6.58g of SO3 and 1.64g of H2O?

If 21.4g of aluminum is reacted w/ 91.3g of Fe2O3, the products will be Al2O3 and iron. What mass of iron will be produced?

can you please explain these in detail so I understand. =]

Answer 1

It is true that sulfur exists as S8, but that's not useful for this problem. So use the equation:

2S + 3O2 ===> 2SO3

Atomic weights: S=32 O=16 O2=32 SO3=80 H2O=18 H2SO4=98

31.5gS x 1molS/32gS = 0.984 mole S

8.65gO2 x 1molO2/32gO2 = 0.270 mole O2

The equation specifies S:O2 at 2:3. You have 0.270 mole O2, which is far less than 0.984, so O2 will run out first. O2 is the limiting reagent.

0.270molO2 x 2molSO3/3molO2 x 80gSO3/1molSO3 = 14.4g SO3

SO3 + H2O ===> H2SO4

6.58gSO3 x 1molSO3/80gSO3 = 0.0822 mole SO3

1.64gH2O x 1molH2O/18gH2O = 0.0911 mole H2O

SO3 is the limiting reagent:

0.0822molSO3 x 1molH2SO4/1molSO3 x 98gH2SO4/1molH2SO4 = 8.06gH2SO4

2Al + Fe2O3 ===> Al2O3 + 2Fe

Atomic weights: Al=27 Fe=56 O=16 Al2O3=102 Fe2O3=160

21.4gAl x 1molAl/27gAl = 0.792 mole Al

91.3gFe2O3 x 1molFe2O3/160gFe2O3 = 0.571molFe2O3

Fe2O3 is the limiting reagent.

0.571molFe2O3 x 2molFe/1molFe2O3 x 56gFe/1molFe = 64.0g Fe

Answer 2

a) mole of S8 = mass / molar mass = 31.5 / 256 = 0.123 moles
mole of O2 = 8.65 / 32 = 0.270 moles

S8 + 12 O2 --> 8 SO3

so, for each mole of S8 used, we need to use 12 moles of O2. if we use all the S8, there won't be enough O2. Thus we use all O2 but only some of the S8. Therefore the counting is based on the moles of O2 (used).

then mole of SO3 produced = 8 / 12 * mole of O2 used = 8 / 12 * 0.270 = 0.180 moles
mass of SO3 = 0.180 * 80 = 14.4 grams

b) mole of SO3 = 6.58 / 80 = 0.082 moles
mole of H2O = 1.64 / 18 = 0.091 moles

SO3 + H2O --> H2SO4

so, for each mole of SO3 used, we need to use 1 mole of H2O. if we use all the H2O, there won't be enough SO3. Thus we use all SO3 but only some of the H2O. Therefore the counting is based on the moles of SO3 (used).

then mole of H2SO4 produced = 1 / 1 * 0.082 = 0.082 moles
mass of H2SO4 = 0.082 * 98 = 8.036 grams

c) moles of Al = 21.4 / 27 = 0.793 moles
moles of Fe2O3 = 91.3 / 160 = 0.571 moles

2 Al + Fe2O3 --> Al2O3 + 2 Fe

so, for each mole of Fe2O3 used, we need to use 2 moles of Al. if we use all the Fe2O3, there won't be enough Al. Thus we use all Al but only some of the Fe2O3. Therefore the counting is based on the moles of Al (used).

moles of Fe = 2 / 2 * moles of Al = 2 / 2 * 0.793 = 0.793 moles
mass = 0.793 * 56 = 44.408 grams