8.4-8.5 Stoichiometry + Limiting Reactant & Percent Yield

Stoichiometry

The quantitative relation between the number of moles (or mass or atoms) of various products and reactants in a chemical reaction.

Atoms and molecules: 4P(s) + 5O2(g)→ 2P2O5(s)

Reactants: 4 molecules of P, 5 molecules of O Products: 2 molecules of diphosophrous pentoxide

4 atoms of P on both, 10 atoms of O on both

In the reactants and products, molecules do/do not have to balance

Do not

In the reactants and products, atoms do/do not have to balance

Do

Atoms and molecules: N2(g) + 3H2(g) → 2NH3(g)

Molecules: 1 N, 3 H → 2 NH

Atoms: 2 N, 6 H → 2 N, 6 H

Moles:

4P(s) + 5O2(g) → 2P2O5(s)

Moles: 4 P, 5 O → 2 Diphosophrous pentaoxide

Mole ratios: 4:5, 5:4, 4:2, 2:4, 5:2, 2:5

Moles:

N2(g) + 3H2(g) → 2NH3(g)

Moles: 1 N, 3 H → 2 NH

Mole ratios: 1:3, 3:1, 1:2, 2:1, 3:2, 2:3

How many miles of hydrogen gas are needed to completely react with 12.3 moles of nitrogen gas?

36.9 mol H2

2NH3→ N2 + 3H2

How many grams of H2 can be produced from 3.5 L of NH3?

0.47 g H2

2H2 + O2 → 2H2O

How many grams of H2O can be produced from 3.5 × 10^²0 molecules of H2?

0.01

2H2 + O2 →2H2O

How many grams of O2 does it take to produce 4.89g of H2O?

4.34g O2

Limiting reagent

The reactant that is totally consumed (used up) in a chemical reaction, limiting the reaction.

Excess reagent

The reactant that is left over (not used up) when the limiting reactant runs out.

N2(g) + 3H2(g) → 2NH3(g)

Lets say you have 2 molecules of N2 and 3 molecules of H2.

Which will you run out of first?

Hydrogen.

N2(g) + 3H2(g) → 2NH3(g)

You react 3.75 moles of nitrogen with 10.4 moles of hydrogen. What is the limiting reagent?

Hydrogen.

N2(g) + 3H2(g) → 2NH3(g)

You react 3.75 moles of nitrogen with 10.4 moles of hydrogen. How much product can you form?

6.93 mol NH3

N2(g) + 3H2(g) → 2NH3(g)

You react 3.75 moles of nitrogen with 10.4 moles of hydrogen.

How much of the excess is remaining?

.28 mol N2

2H2 + O2 → 2H2O

What is the limiting reagent and the maximum amount of product formed when 1.25g of H2 reacts with 0.75g of O2?

O2 & 0.85g

2H2 + O2 → 2H2O

What is the amount of excess left over when 1.25g of H2 reacts with 0.75g of O2?

1.16g H2

Theoretical yield

The maximum amount of product that can be produced from a given amount of reacts.

Actual yield

Measured amount of a product obtained from a reaction is called the actual yield of that product.

Percent yield

The ratio of the actual yield to the theoretical yield × 100.

% yield = actual yield (A)/theoretical yield (T) × 100

N2(g) + 3H2(g) → 2NH3(g)

You react excess hydrogen gas with 3.4 moles of nitrogen gas. What is the theoretical yield? If you produced only 5.0 moles of NH3, what was your percent yield?

Theoretical: 6.8mol NH3

Percent: 73.5

N2(g) + 3H2(g) → 2NH3(g)

You react nitrogen gas with hydrogen gas in a process that has an 85% yield. If you start with 5.7 mol H2, how much NH3 did you actually produce?

3.23 mol

2H2 + O2 → 2H2O

What is the percent yield of 0.65g of O2 reacts with excess H2 to produce 0.53g of H2O?

72.4%