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  1. Problems
  2. /Sulfur dioxide, 160 g with a molar mass of 64.0 g/mol, passes over vanadium(V) oxide, and 90.0 % of…
MediumChemistryHigh School (11–12)The d- and f-Block Elements › Transition Elements (d-Block)Numerical

Sulfur dioxide, 160 g with a molar mass of 64.0 g/mol, passes over vanadium(V) oxide, and 90.0 % of…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Sulfur dioxide, 160 g160\ \text{g}160 g with a molar mass of 64.0 g/mol64.0\ \text{g/mol}64.0 g/mol, passes over vanadium(V) oxide, and 90.0 %90.0\ \%90.0 % of it is oxidised: 2SO2+O2→2SO32\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_32SO2​+O2​→2SO3​. Each SO2\text{SO}_2SO2​ gives 2 electrons to the catalyst. Find the moles of SO3\text{SO}_3SO3​ formed, the mass of oxygen used (molar mass 32.0 g/mol32.0\ \text{g/mol}32.0 g/mol) and the moles of electrons that pass through the catalyst.

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Need a reminder? The key ideaShowHide

Moles = mass in g divided by molar mass in g/mol, or concentration in mol/L times volume in litres. Take every mole ratio from the balanced equation, and follow the chain one step at a time.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Moles of SO2\text{SO}_2SO2​. 160 g64.0 g/mol=2.50 mol\dfrac{160\ \text{g}}{64.0\ \text{g/mol}} = 2.50\ \text{mol}64.0 g/mol160 g​=2.50 mol.

Step 2. SO3\text{SO}_3SO3​. 0.900×2.50=2.25 mol0.900 \times 2.50 = 2.25\ \text{mol}0.900×2.50=2.25 mol.

Step 3. Oxygen. The ratio is 2:12 : 12:1, so 1.125 mol1.125\ \text{mol}1.125 mol of O2\text{O}_2O2​. The mass is 1.125×32.0=36.0 g1.125 \times 32.0 = 36.0\ \text{g}1.125×32.0=36.0 g.

Step 4. Electrons. 2×2.25=4.50 mol2 \times 2.25 = 4.50\ \text{mol}2×2.25=4.50 mol.

Answer: 2.25 mol2.25\ \text{mol}2.25 mol, 36.0 g36.0\ \text{g}36.0 g and 4.50 mol4.50\ \text{mol}4.50 mol.

Watch out. The catalyst is not used up. Its mass is the same after the run, because it gives the electrons to the oxygen as soon as it takes them.

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