Question

A. The formation of organic compounds by the reaction (II) sulfide and carbonic acid is described by the following equaition: 2 FeS + H_2CO_3 ---> 2 FeO + 1/m (CHOH)_n + 2 S. How much FeO is produced starting with 1.50 g FeS and 0.515 moles of H_2CO_3 if the reaction results in a 75.50% yield?

B. A sealed chamber contains 7.50 g CH_4 and 12.00 g O_2. The mixture is ignited. How many grams of CO_2 are produced?

Answer 1

a)

moles FeS = 1.50 g /87.913 g/mol=0.0170= limiting reactant

theoretical moles FeO = 0.0170

theoretical mass FeO = 0.0170 mol x 71.846 g/mol= 1.221g

mass FeO produced = 1.221x 75.50/ 100 = 0.9218 g

(or)

The molecular weight of FeS = 55.85 for Fe + 32.06 for S = 87.91
So one mole of FeS has a mass of 87.91 g.
Since we start with 1.50 g of FeS, that corresponds to (1.50 g of FeS)(1 mole / 87.91 g of FeS) = 0.01706 moles of FeS.

So we start with 0.01706 moles of FeS and 0.515 moles of H2CO3. From the balanced equation you can see that for every one mole of H2CO3 that 2 moles of FeS are needed to form 2 moles of FeO.
So FeS is the restrictive reactant. Since we have 0.01706 moles of FeS, then only 0.01706 / 2 = 0.00853 moles of H2CO3 are used up to form 0.01706 moles of FeO.

Now that corresponds to 100% yield, which means the reaction goes according to what it says.
75.50% yield that only 75.50% of that amount is actaully formed. So only (0.75)(0.01706 moles) = 0.01279 moles of FeO are formed. This corresponds to
(0.01279moles of FeO)(71.84 g/mole of FeO) = 0.92g. of FeO

b.

CH4 + 2O2 = CO2 + 2 H2O
given:
CH4 = 7.50g
O2 = 12.00g
1mol CH4 = 75g/mol
1mol O2 = 63.38g/mol
1mol CO2 = 44.01g/mol
7.504g CH4 * 1mol/75g CH4 = 0.1mol
12.00g O2 * 1mol/63.98g = 0.1063mol
limiting reactant is CH4
so,
44.01g/mol CO2 * 0.1mol CH4 = 4.401g CO2