Answer:
5.38 L
Option D.
Explanation:
2 NaOH(s) + 2 Al(s) + 6 H₂O(l) → 2 NaAl(OH)₄(s) + 3 H₂(g)
We convert mass of Al to moles:
4.32 g . 1 mol /26.98g = 0.160 moles
As NaOH is in excess, aluminum is the limiting reactant.
We see stoichiometry, were ratio is 2:3.
2 moles of Al can produce 3 moles of hydrogen
Our 0.160 moles may produce (0.160 . 3)/2 = 0.240 moles of H₂.
We know that 1 mol of any gas at STP conditions is contained in 22.4L
So let's make the conversion factor:
0.240 mol . 22.4L / 1mol = 5.38 L
