Chemical Calculations
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Step 1: Write balanced equation for reaction.
Step 2: Calculate amount of reactants:
| Data given: | Formula |
| Number of particles (N) | N/L L: Avogadro’s number |
| Mass of substance (m) | m/M M: molar mass |
| Volume of gas (V) | V/Vm Vm: molar volume (22.4 dm3 at stp; 24.0 dm3 at rtp) |
| Concentration of solution (c) | cV V: volume of solution |
Step 3: Determine limiting reagent.
Step 4: Use reacting ratio to obtain amount of product based on limiting reagent.
| Percentage yield | = | actual mass of product |
| theoretical (calculated) mass of product |
| Percentage purity | = | mass of substance |
| mass of mixture |
| Example: 5.00 g of aluminium and 5.00 g of sulfur are ignited in a furnace to form aluminium sulfide. What is the maximum mass of product that can be formed?
Step 1: Write balanced equation for reaction. 2Al + 3S → Al2S3
Step 2: Calculate amount of reactants. n(Al) = 5.00/26.98 = 0.185 mol n(S) = 5.00/32.1 = 0.156 mol
Step 3: Determine limiting reagent. 1 mol Al reacts with 1.5 mol S → 0.185 mol Al reacts with 0.278 mol S → S is limiting
Step 4: Use reacting ratio to obtain amount of product based on limiting reagent. n(Al2S3) = 1/3 x 0.156 = 0.0519 mol m(Al2S3) = n(Al2S3) x Mr(Al2S3) = 0.0519 x 150.26 = 7.80 g |
Have problems identifying which reagent is limiting?
Use the amount of each reactant to find the amount of product based on reacting ratio. The reactant which gives the lesser amount is limiting.
| Example: (from qns above) 2Al + 3S → Al2S3 n(Al) = 5.00/26.98 = 0.185 mol n(S) = 5.00/32.1 = 0.156 mol
If Al is limiting → n(Al2S3) = 1/2 x 0.185 = 0.0927 mol If S is limiting → n(Al2S3) = 1/3 x 0.156 = 0.0519 mol (< 0.0927) → S is limiting |
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