Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) When 25.0 g of Zn reacts, how many L of H2 gas are formed at STP?

Chemistry

2 answers:

ludmilkaskok [199]3 years ago

7 0

Answer to this is 8.56L

irga5000 [103]3 years ago

4 0

Answer : The volume of hydrogen gas formed at STP are, 8.557 liters

Explanation : Given,

Mass of zinc = 25.0 g

Molar mass of zinc = 65.38 g/mole

First we have to calculate the moles of zinc.

$\text{Moles of }Zn=\frac{\text{Mass of }Zn}{\text{Molar mass of }Zn}=\frac{25g}{65.38g/mole}=0.382moles$

Now we have to calculate the moles of hydrogen gas.

The given balanced chemical reaction is,

$Zn(s)+2HCl(aq)\rightarrow H_2(g)+ZnCl_2(aq)$

From the balanced chemical reaction, we conclude that

As, 1 mole of Zn react to give 1 mole of hydrogen gas

So, 0.382 mole of Zn react to give 0.382 mole of hydrogen gas

Now we have to calculate the volume of hydrogen gas.

At STP,

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

So, 0.382 mole of hydrogen gas contains $0.382\times 22.4=8.557L$ volume of hydrogen gas.

Therefore, the volume of hydrogen gas formed at STP are, 8.557 liters

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