Question
Asked Oct 2, 2019

Zinc reacts with hydrochloric acid according to the reaction equation

 

Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)

 

How many milliliters of 5.50 M HCl(aq)5.50 M HCl(aq) are required to react with 2.05 g Zn(s)?

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Step 1

The reaction of zinc with hydrochloric acid is shown as,

Zn(s)2HCI (aq)
ZnCl2 (aq) H2 (g)
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Zn(s)2HCI (aq) ZnCl2 (aq) H2 (g)

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Step 2

It is required to calculate the volume in mL of 5.50 M HCl which will react with 2.05 g of Zinc. To calculate the volume, first the number of moles of HCl must be calculated, which can be done as,

Mass
No. of moles off Zinc =Molar Mass
2.05 g
No. of moles of Zinc 65.38 g/mol
No. of moles of Zinc 0.0313
For 1 mole of zinc 2 moles of HCl is required,
Therefore, moles of HCl = 2 x 0.0313 0.0627
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Mass No. of moles off Zinc =Molar Mass 2.05 g No. of moles of Zinc 65.38 g/mol No. of moles of Zinc 0.0313 For 1 mole of zinc 2 moles of HCl is required, Therefore, moles of HCl = 2 x 0.0313 0.0627

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Step 3

Now the volume of 5.5 M HCl ...

Molarity=No. of moles of solute
Volume (L)
0.0627
5.5 Volume (L)
Volume 0.0114 L
Since 1 L 1000 mL
Hence the volume = 11.40 mL
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Molarity=No. of moles of solute Volume (L) 0.0627 5.5 Volume (L) Volume 0.0114 L Since 1 L 1000 mL Hence the volume = 11.40 mL

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