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The action of some commercial drain cleaners is based on the following reaction:
2 NaOH(s) + 2 Al(s) + 6 H2O(l) â 2 NaAl(OH)4(s) + 3 H2(g)
What is the volume of H2 gas formed at STP when 4.32 g of Al reacts with excess NaOH?
A. 3.59 L
B. 2.39 L
C. 5.87 L
D. 5.38 L

Respuesta :

anfabba15

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

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