Content Review Week 3

Conversions

100

How many meters are in 23 km?

23km x 1000m/1km= 23000 or 2.3x10^4m

200

How many grams are in 12 mg?

12mg x 1g/1000mg= 0.012 or 1.2x10^-2g

300

How many microliters are in 420. L?

420.L x (1x10^6𝜇𝐿)/ 1L= 4.2x10^8𝜇𝐿

400

How many centimeters are in 300. μm?

300. μm x (1m/10^6μm) x (100cm/1m)= 3x10^-2cm

500

Children’s Chewable Advil contains 100. mg of ibuprofen per tablet. If the recommended dosage is 10. mg/kg, how many tablets are needed for a 37 lb child? Round your answer to the nearest whole tablet. (1 kg = 2.2046 lb)

 

37 lb x (1 kg/2.2046 lb) x (10mg/1 kg) x (1 tablet/100mg)= 1.678 tablets= 2 tablets

Theoretical Yield

100

You can make one sandwich from 2 slices of bread and 1 slice of cheese. You have 10 slices of bread and 3 slices of cheese.

Limiting ingredient: cheese

• Maximum sandwiches: 3

• Bread left over: 10 − 2×3 = 4 slices

200

One pizza requires 3 cups of flour and 2 cups of cheese. You have 15 cups of flour and 8 cups of cheese.

• Flour allows: 15/3 = 5 pizzas

• Cheese allows: 8/2 = 4 pizzas

• Limiting ingredient: cheese

• Maximum pizzas: 4

• Flour left: 15 - 4×3 = 3 (1 pizza)

300

Nitrogen and hydrogen forming ammonia

Reaction: N₂ + 3H₂ → 2NH₃

Given: 1.0 mol N₂ and 2.0 mol H₂.

• Stoichiometry: 1 N₂ : 3 H₂

• N₂ needs H₂: 1.0 × 3 = 3.0 mol H₂ (only 2.0 available)

• Limiting reactant: H₂

• NH₃ produced: 2.0 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 1.33

mol NH₃

• Leftover N₂: 1.0 − 0.667 ≈ 0.33 mol

400

Combustion of propane

Reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Given: 0.80 mol C₃H₈ and 3.0 mol O₂.

 Stoichiometry: 1 C₃H₈ : 5 O₂

• C₃H₈ needs O₂: 0.80 × 5 = 4.0 mol O₂ (only 3.0 available)

• Limiting reactant: O₂

• CO₂ produced: 3.0 mol O₂ × (3 mol CO₂ / 5 mol O₂) = 1.8

mol CO₂

• Leftover C₃H₈: 0.80 − 0.60 = 0.20 mol

500

Neutralization (sulfuric acid and sodium hydroxide)

Reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Given: 0.30 mol H₂SO₄ and 0.40 mol NaOH

 Stoichiometry 1 H₂SO₄ : 2 NaOH

• H₂SO₄ would need: 0.30 × 2 = 0.60 mol NaOH

(only 0.40 present)

• Limiting reactant: NaOH

• Na₂SO₄ produced: 0.40 mol NaOH × (1 mol Na₂SO₄ / 2 mol

NaOH) = 0.20 mol Na₂SO₄

• H₂SO₄ used: 0.40 mol NaOH × (1 mol H₂SO₄ / 2 mol

NaOH) = 0.20 mol H₂SO₄

• Leftover H₂SO₄: 0.30 − 0.20 = 0.10 mol

Balancing Equations

100

H₂ + O₂ → H₂O

2H₂ + O₂ → 2H₂O

200

N₂ + H₂ → NH₃

N₂ + 3H₂ → 2NH₃

300

Fe + O₂ → Fe₂O₃

4Fe + 3O₂ → 2Fe₂O

400

C₂H₆ + O₂ → CO₂ + H₂O

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

500

C₄H₁₀ + O₂ → CO₂ + H₂O

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Naming Compounds

100

Fe2O3

iron(III) oxide

200

lead(IV) oxide

PbO2

300

calcium nitrate

Ca(NO3)2

400

ammonium carbonate

(NH4)2CO3

500

ammonium hypochlorite

NH4ClO

Sig Figs

100

125.67+3.4−12.089

117.1

200

4.56×0.00320

0.0146

300

(45.67-2.3)/6.20

7.00

400

(2.50×10^3)(4.62×10^−2)

1.16×10^2

500

((125.6−17.23)(3.40))/(2.5+0.37)

1.3x10^2