Bev93
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A 6.72 g sample of zinc was placed in 100.0 mL of 1.50 mol/L hydrochloric acid. After all reaction stops, how much zinc should remain?

Sagot :

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Answer:

1.82g of Zn remain

Explanation:

Zinc, Zn, reacts with hydrochloric acid, HCl, as follows:

Zn + 2HCl → ZnCl2 + H2

Where 1 mole of Zn reacts with 2 moles of HCl

To solve this question we must find the moles of each reactant. With the moles of HCl we can find the moles of zinc needed for a complete reaction and the mass of zinc that remains:

Moles Zn -Molar mass: 65.32g/mol-

6.72g * (1mol / 65.38g) = 0.1028 moles Zn

Moles HCl:

100.0mL = 0.1000L * (1.50mol / L) = 0.150moles HCl

The moles of Zn that react with 0.150 moles HCl are:

0.150 mol HCl * (1mol Zn / 2mol HCl) = 0.075 moles of Zn are required

The moles that remain are:

0.1028 moles Zn - 0.075 moles of Zn = 0.0278 moles Zn

The mass is:

0.0278 moles Zn * (65.38g / mol) =

1.82g of Zn remain

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