Guided example: solution

As a recipe: 8.0 mol Fe is twice the 4 mol Fe in the equation, so it takes twice the 3 mol O₂, or 6.0 mol.

Guided example: solution

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer A

B read O₂'s coefficient alone: 7 mol O₂ pairs with 2 mol C₂H₆, not 1. C inverted the ratio: 1 × (2/7) = 0.286. D answered with C₂H₆'s coefficient, 2, without applying the ratio.

Worked example 1: solution

One conversion factor is needed.

Step 2 · Build the ratio, wanted over given

Two orientations exist. Only one cancels the given unit:

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Practice 2: answer B

A assumed 1:1: 7.00 × 1 = 7.00. C inverted the ratio: 7.00 × (1/2) = 3.50. D answered with the coefficient, 2.

Practice 3: answer D

A assumed 1:1: 0.52. B inverted the ratio: 0.52 × (2/15) = 0.069. C read CO₂'s coefficient instead of O₂'s: 0.52 × (12/2) = 3.1.

Worked example 2: solution

One conversion factor is needed.

Step 2 · Build the ratio, wanted over given

The ratio, wanted on top: 2 mol KClO₃ over 3 mol O₂, so mol O₂ cancels.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 4: answer A

B assumed 1:1: 0.250. C divided by CO₂'s 2 but left out H₂O's 3: 0.250 × (1/2) = 0.125. D inverted the ratio: 0.250 × (2/3) = 0.167.

Practice 5: answer C

A assumed a 1 : 1 ratio: 26.0 × 1 = 26.0. B multiplied by 2 without dividing by 8: 26.0 × 2 = 52.0. D inverted the ratio: 26.0 × (8/2) = 104.

Practice 6: answer D

Practice 6: answer D

A stopped at the CO₂ alone: 1.50. B assumed 1:1 for each product: 2.40 + 2.40 = 4.80. C inverted both ratios: 2.40 × (8/5) + 2.40 × (8/6) = 3.84 + 3.20 = 7.04.

Worked example 1: solution

Two conversion factors are needed.

Step 1 · Grams → moles

CH₄'s own molar mass converts the given mass to moles:

Worked example 1: solution

The mole ratio, written CO₂ over CH₄ (1 : 1), crosses substances:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 1: one more factor

Step 3 · Moles → grams

CO₂'s molar mass converts the moles out to mass. The same chain, one factor longer:

Worked example 1: one more factor

Step 3 · Moles → grams

CO₂'s molar mass converts the moles out to mass. The same chain, one factor longer:

Worked example 1: the grams route on the map

Worked example 2: solution

Three conversion factors are needed.

Step 1 · Grams → moles

Glucose's own molar mass converts the given mass to moles. Its unit cancels the given unit:

Worked example 2: solution

The mole ratio is the only factor that crosses substances. Two orientations exist. Only one cancels mol glucose:

Worked example 2: solution

Convert out with the target's molar mass, in one continuous setup. Carry all digits and round once at the end:

Worked example 2: solution

Worked example 2: the route on the map

Practice 1: answer C

A skipped the mole ratio: 25.0/44.09 × 18.02 = 10.2. B stopped at moles: 25.0/44.09 × 4 = 2.27 mol H₂O, one factor short of grams. D inverted the ratio: 25.0/44.09 × (1/4) × 18.02 = 2.55.

Practice 2: answer B

A stopped at moles: 35.0/26.98 × (2/4) = 0.649 mol Al₂O₃, one factor short of grams. C skipped the mole ratio: 35.0/26.98 × 101.96 = 132. D inverted the ratio: 35.0/26.98 × (4/2) × 101.96 = 265.

Practice 3: answer D

A skipped the mole ratio: 2.20/44.01 × 84.01 = 4.20. B inverted the ratio: 2.20/44.01 × (1/2) × 84.01 = 2.10. C stopped at moles: 2.20/44.01 × 2 = 0.100 mol NaHCO₃, one factor short of grams.

Worked example 3: solution

Three conversion factors are needed.

Step 1 · Grams → moles

NaN₃'s molar mass, 65.02 g per mole, converts 50.0 g to 0.769 mol NaN₃.

Worked example 3: solution

The mole ratio, written N₂ over NaN₃ (3 : 2), crosses substances and turns 0.769 mol NaN₃ into 1.1535 mol N₂, unrounded.

Worked example 3: solution

Avogadro's number converts out, with mol N₂ on the bottom so it cancels. Carry all digits and round once at the end:

Worked example 3: solution

Run backward, the chain starts at a count: molecules B → mol B → mol A → g A, with Avogadro's number flipped so molecules cancel.

Worked example 3: the route on the map

Practice 4: answer B

A stopped at moles: 5.00/114.22 × (16/2) = 0.350 mol CO₂, one factor short of a count. C flipped Avogadro's number: 0.350 ÷ 6.022 × 10²³ = 5.82 × 10⁻²⁵, less than one molecule. D skipped the mole ratio: 5.00/114.22 × 6.022 × 10²³ = 2.64 × 10²².

Worked example 1: which reactant limits

Step 1 · Convert both reactants to moles

Both amounts are already in moles, so Step 1 is complete.

Step 2 · Divide each by its coefficient

Worked example 1: which reactant limits

Worked example 1: which reactant limits

Worked example 1: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 1: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 2: which reactant limits

Four conversion factors are needed: one molar mass for each reactant, then the mole ratio, then the target's molar mass.

Step 1 · Convert both reactants to moles

Worked example 2: which reactant limits

Worked example 2: which reactant limits

Worked example 2: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 2: how much product

Step 4 · Compute product from the limiting reactant only

Worked example 2: the route on the map

Worked example 3: the NH₃ each reactant can make

Three conversion factors per reactant: its molar mass, the mole ratio, and the molar mass of NH₃.

Step 1 · Grams to moles, for each reactant

Worked example 3: the NH₃ each reactant can make

Worked example 3: the NH₃ each reactant can make

Worked example 3: grams of NH₃, then compare

Step 3 · Moles to grams of product

Worked example 3: grams of NH₃, then compare

Step 3 · Moles to grams of product

Worked example 3: the route on the map

Worked example 3: both paths agree

Worked example 3: both paths agree

Worked example 3: needed vs loaded

A third check: amount needed vs amount loaded

Convert one reactant into the amount of the other that it needs:

Worked example 3: needed vs loaded

A third check: amount needed vs amount loaded

Convert one reactant into the amount of the other that it needs:

Practice 1: answer A

B stopped at moles: 30.0/70.90 × (2/3) = 0.282 mol AlCl₃, one factor short of grams. C skipped the mole ratio: 30.0/70.90 × 133.33 = 56.4. D inverted the ratio: 30.0/70.90 × (3/2) × 133.33 = 84.6.

Practice 2: answer B

A took the smaller mass, H₂, as limiting: 3.00/2.016 × (1/2) × 32.04 = 23.8. C skipped the grams-to-moles step and treated grams of CO as moles: 14.0 × 32.04 = 449. D stopped at moles: 14.0/28.01 × (1/1) = 0.500 mol CH₃OH, one factor short of grams.

Practice 3: answer C

Practice 3: answer C

A kept the larger amount, from CO₂, the excess (also the smaller mass and fewer moles): 53.0. B stopped at moles: 0.900 × (1/2) = 0.450. D inverted the ratio: 0.900 × 2 × 105.99 = 191. E skipped the ratio: 0.900 × 105.99 = 95.4.

Practice 5: answer D

Practice 5: answer D

A stopped at the Mg consumed, 12.82 g. B inverted the ratio: 20.00 − 50.00/189.67 × (1/2) × 24.31 = 16.80. C skipped it: 20.00 − 50.00/189.67 × 24.31 = 13.59.

Worked example 1: solution

Three conversion factors build the theoretical yield.

Step 1 · Compute the theoretical yield

Worked example 1: solution

The balance reads 18.5 g. The first attempt divided by 50.0 g of CaCO₃, a different substance. The 100% mark is 22.0 g: the most CO₂ this charge can form.

Worked example 1: solution

Worked example 1: solution

Worked example 2: which reactant limits

Four conversion factors are needed: one molar mass for each reactant, then the mole ratio and the product's molar mass.

Step 1 · Compute the theoretical yield

Worked example 2: which reactant limits

Worked example 2: which reactant limits

Worked example 2: the percent yield

Step 1 · Compute the theoretical yield

Worked example 2: the percent yield

Step 1 · Compute the theoretical yield

The dried precipitate weighs 14.2 g. Measured on the balance, not computed.

Worked example 2: the percent yield

Step 1 · Compute the theoretical yield

Worked example 2: the percent yield

Step 1 · Compute the theoretical yield

Practice 1: answer B

A flipped the fraction, theoretical over actual: 18.81/15.6 × 100 = 121%, more product than the reaction can make. C left the fraction as a decimal: 15.6/18.81 = 0.829, and the × 100 makes it 82.9%. D skipped the mole ratio in the theoretical yield: 40.0/34.02 × 32.00 = 37.6 g, then 15.6/37.6 × 100 = 41.5%.

Practice 1: answer B

Practice 2: answer D

Practice 2: answer D

Practice 2: answer D

Practice 3: answer A

B stopped at the theoretical yield: 5.20 g is the 100% mark, not what the lamp delivers. C divided by the yield instead of multiplying: 5.20 ÷ 0.800 = 6.50, more than the maximum. D applied the yield to the starting material: 12.8 × 0.800 = 10.2.

Practice 3: answer A