Question

d. if you have 80.0 grams of lye (naoh), how many moles would you have? show your work (2pts: 1pt for a correct answer, 1pt for showing work)
e. if you have the number of moles of naoh above, how many moles of h₂ would also be produced? (1pts: 0.5 pt for correct answer, 0.5 pt for showing work)
f. setup the full stoichiometric equation for the reaction to calculate how many grams of hydrogen gas (h₂) would be produced if 80.0 grams of lye (naoh) were also produced (3 pts: 1 pt for correct answer, 1 point for correct units, 1 point for proper set up)

Explanation:

Step1: Calculate molar mass of NaOH

Molar mass of $\text{NaOH} = 23.0 + 16.0 + 1.0 = 40.0\ \text{g/mol}$

Step2: Find moles of NaOH (part d)

$\text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{80.0\ \text{g}}{40.0\ \text{g/mol}} = 2.00\ \text{mol}$

Step3: Use reaction stoichiometry (part e)

The balanced reaction for NaOH and H₂ production is: $2\text{Na} + 2\text{H}_2\text{O}
ightarrow 2\text{NaOH} + \text{H}_2$
Mole ratio $\text{H}_2:\text{NaOH} = 1:2$
$\text{Moles of H}_2 = 2.00\ \text{mol NaOH} \times \frac{1\ \text{mol H}_2}{2\ \text{mol NaOH}} = 1.00\ \text{mol}$

Step4: Calculate mass of H₂ (part f)

Molar mass of $\text{H}_2 = 2.0\ \text{g/mol}$
$\text{Mass of H}_2 = 1.00\ \text{mol} \times 2.0\ \text{g/mol} = 2.00\ \text{g}$

Answer:

d. 2.00 moles
e. 1.00 moles
f. 2.00 grams