Question

The action of some commercial drain cleaners is based on the following reaction: 2 NaOH(s) + 2 Al(s) + 6 H2O(l) → 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? The action of some commercial drain cleaners is based on the following reaction: 2 NaOH(s) + 2 Al(s) + 6 H2O(l) → 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? 3.59 L 2.39 L 5.87 L 5.38 L

Answer 1

The action of some commercial drain cleaners is based on the following reaction: 2 NaOH(s) + 2 Al(s) + 6 H2O(l) → 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?

3.59 L

2.39L

5.87 L

5.38 L---answer

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explanation

Given data that

T= 273.15K
P=0.986atm
2 NaOH(s) + 2 Al(s) + 6 H2O(l) → 2 NaAl(OH)4(s) + 3 H2(g)

because of STP

aluminium

gram-molecular mass = 26.98g/mol,

you can find that there are 0.1601 moles.

Since Al is in a 2:3 ratio with H2 and there is an excess of NaOH,

4.32 /26.98g =0.1601 moles Of aluminium

4.32 g of Al reacts with excess

equation PV=nRT

2 moles Al(s) ---------------------------3 moles 3 H2(g)

end with 0.2415 moles of H2.

Then, use the equation PV=nRT to get your answer.

(0.986atm)V=(.2415mol)(0.0821)(273.15K...

V= (.245*.0821*273.15)/.986 = 5.5723 L