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The following reaction is done at
T = 25°C and P= 1.0 atm:

Ca (s) + 2 HCl (aq) → CaCl_2 (aq) + H_2 (g)

If 500.0 g of calcium are added to an excess of HCI, what volume of H_2 is produced?​

The following reaction is done at T = 25°C and P= 1.0 atm:Ca (s) + 2 HCl (aq) → CaCl2 (aq) + H2 (g)If 500.0 g of calcium are added to an excess of HCI, what vol class=

Answer :

Eduard22sly

Answer: 280dm3

Explanation:Please see attachment for explanation

${teks-lihat-gambar} Eduard22sly

Answer : The volume of [tex]H_[/tex] gas produced is, 305.8 L

Explanation :

First we have to calculate the moles of Ca.

[tex]\text{Moles of Ca}=\frac{\text{Mass of Ca}}{\text{Molar mass of Ca}}[/tex]

Molar mass of Ca = 40 g/mol

[tex]\text{Moles of Ca}=\frac{500.0g}{40g/mol}=12.5mol[/tex]

Now we have to calculate the moles of [tex]H_2[/tex] gas.

The balanced chemical reaction is:

[tex]Ca(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2(g)[/tex]

From the balanced chemical reaction we conclude that,

As, 1 mole of Ca react to give 1 mole of [tex]H_2[/tex] gas

So, 12.5 mole of Ca react to give 12.5 mole of [tex]H_2[/tex] gas

Now we have to calculate the volume of [tex]H_[/tex] gas.

Using ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]H_2[/tex] gas = 1.0 atm

V = Volume of [tex]H_2[/tex] gas = ?

n = number of moles [tex]H_2[/tex] = 12.5 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]H_2[/tex] gas = [tex]25^oC=273+25=298K[/tex]

Putting values in above equation, we get:

[tex]1.0atm\times V=12.5mole\times (0.0821L.atm/mol.K)\times 298K[/tex]

[tex]V=305.8L[/tex]

Thus, the volume of [tex]H_[/tex] gas produced is, 305.8 L

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