Question

Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g) What t mass of hydrogen gas would be formed from a reaction of 1.83 g Al and 4.30 g NaOH in water?

Answer 1

Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

Step 1: Data given

Mass of 1.83 grams of Al

Mass of NaOH = 4.30 grams

Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

Step 2: The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

Step 3: Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

Step 4: Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

Step 5: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

Step 6: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

Step 7: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams