Answer:
The mass of hydrogen gas formed is 0.205 grams
Explanation:
<u>Step 1:</u> 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
<u>Step 2:</u> The balanced equation:
2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)
<u>Step 3:</u> 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
<u>Step 4:</u> Calculate moles of NaOH
Moles NaOH = 4.30 grams / 40 g/mol
Moles NaOH = 0.1075 moles
<u>Step 5</u>: 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
<u>Step 6</u>: 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
<u>Step 7</u>: 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
