Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 4.504.50 M HCl(aq) are required to react with 7.557.55 g of an ore containing 49.049.0% Zn(s) by mass?volume: mL

Question
Asked Nov 6, 2019

Zinc reacts with hydrochloric acid according to the reaction equation shown.

 

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

 

How many milliliters of 4.504.50 M HCl(aq) are required to react with 7.557.55 g of an ore containing 49.049.0% Zn(s) by mass?

volume:
 
mL
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Step 1

The gram of Zn can be calculated as

49.049
3.7031g of Zn(S)
7.55g of an Orex.
100
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49.049 3.7031g of Zn(S) 7.55g of an Orex. 100

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Step 2

The reaction of zinc with hydrochloric acid is shown as,

Zn(s)+2HC1(aq) ZnCl, (aq)+H2 (g)
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Zn(s)+2HC1(aq) ZnCl, (aq)+H2 (g)

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Step 3

It is required to calculate the volume in mL of 4.50 M HCl which will react with 3.7031 g of Zinc. To c...

Mass
No.of moles of Zn=-
Molar mass
3.7031g
No.of moles of Zn=-
65.36g/mol
No.of moles of Zn 0.05665 mole
For1mole of Zn 2 moles of HCl isrequired
Hence,the moles of HC=2x0.05665-0.1133 moles.
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Mass No.of moles of Zn=- Molar mass 3.7031g No.of moles of Zn=- 65.36g/mol No.of moles of Zn 0.05665 mole For1mole of Zn 2 moles of HCl isrequired Hence,the moles of HC=2x0.05665-0.1133 moles.

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