Question

chem 20 - stoichiometry check in (d1/d2a) nov 7th
please show your work and make it obvious which number is your final answer.
2k(s) + 2h₂o(l) → 2koh(aq) + h₂(g)
a) using the reaction equation above - how many mols of hydrogen would be produced if 2.5 moles of potassium is added to an excess amount of water? (2 marks)
b) a teacher tries to impress her students by adding 42.5 g of potassium to 45.00 g of water.
how much gaseous product will form? (8 marks)
c) how much leftover reactant will you have? (2 marks)
d) if 0.8320 g of hydrogen gas is produced in the lab, what is the percent yield of this reaction?
(if you were unable to answer q2b, use 1.50 g as your theoretical value) (2 marks)

Explanation:

Part (a)

Step1: Identify mole ratio

From the reaction \( \ce{2 K_{(s)} + 2 H_2O_{(l)} -> 2 KOH_{(aq)} + H_2_{(g)}} \), the mole ratio of \( \ce{K} \) to \( \ce{H_2} \) is \( 2:1 \).

Step2: Calculate moles of \( \ce{H_2} \)

Given moles of \( \ce{K} = 2.5 \) mol. Let moles of \( \ce{H_2} = x \). Using the ratio \( \frac{2}{1}=\frac{2.5}{x} \), solve for \( x \): \( x=\frac{2.5\times1}{2}=1.25 \) mol.

Step1: Calculate moles of reactants

Molar mass of \( \ce{K} = 39.10 \) g/mol, moles of \( \ce{K} = \frac{42.5\ \text{g}}{39.10\ \text{g/mol}} \approx 1.087 \) mol.
Molar mass of \( \ce{H_2O} = 18.02 \) g/mol, moles of \( \ce{H_2O} = \frac{45.00\ \text{g}}{18.02\ \text{g/mol}} \approx 2.497 \) mol.

Step2: Determine limiting reactant

From the reaction, mole ratio of \( \ce{K}:\ce{H_2O} = 2:2 = 1:1 \).
Moles of \( \ce{K} \) (1.087 mol) < Moles of \( \ce{H_2O} \) (2.497 mol), so \( \ce{K} \) is limiting.

Step3: Calculate moles of \( \ce{H_2} \) from limiting reactant

Mole ratio of \( \ce{K}:\ce{H_2} = 2:1 \). Moles of \( \ce{H_2} = \frac{1.087\ \text{mol}}{2} \approx 0.5435 \) mol.

Step4: Calculate mass of \( \ce{H_2} \)

Molar mass of \( \ce{H_2} = 2.02 \) g/mol, mass of \( \ce{H_2} = 0.5435\ \text{mol} \times 2.02\ \text{g/mol} \approx 1.10\ \text{g} \) (or using the alternate 1.50 g if part (b) was not solved, but here we use calculated).

Step1: Moles of \( \ce{H_2O} \) reacted

Since \( \ce{K} \) is limiting, moles of \( \ce{H_2O} \) reacted = moles of \( \ce{K} = 1.087 \) mol.

Step2: Moles of \( \ce{H_2O} \) leftover

Moles of \( \ce{H_2O} \) leftover = initial moles - reacted moles = \( 2.497 - 1.087 = 1.410 \) mol.

Step3: Mass of \( \ce{H_2O} \) leftover

Mass = \( 1.410\ \text{mol} \times 18.02\ \text{g/mol} \approx 25.41 \) g.

Answer:

1.25 moles

Part (b)