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Damm [24]
3 years ago
7

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 6.32 g of Al reacts with excess NaOH?
Chemistry
1 answer:
Blababa [14]3 years ago
6 0

Answer : The volume of H_2 gas formed at STP is 7.86 liters.

Explanation :

The balanced chemical reaction will be:

2NaOH(s)+2Al(s)+6H_2O(l)\rightarrow 2AnAl(OH)_4(s)+3H_2(g)

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}

Molar mass of Al = 27 g/mole

\text{Moles of }Al=\frac{6.32g}{27g/mole}=0.234mole

Now we have to calculate the moles of H_2 gas.

From the reaction we conclude that,

As, 2 mole of Al react to give 3 mole of H_2

So, 0.234 moles of Al react to give \frac{0.234}{2}\times 3=0.351 moles of H_2

Now we have to calculate the volume of H_2 gas formed at STP.

As, 1 mole of H_2 gas contains 22.4 L volume of H_2  gas

So, 0.351 mole of H_2 gas contains 0.351\times 22.4=7.86L volume of H_2 gas

Therefore, the volume of H_2 gas formed at STP is 7.86 liters.

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Answer:

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