Question

Zinc reacts with hydrochloric acid according to the reaction equation Zn (s)+2HCl (aq)--> ZnCl2 (aq) + H2(g) How many milliliters of 2.50 M HCl(aq) are required to react with 3.55 g of an ore containing 37.0% Zn(s) by mass?

Answer #1

Zn (s) + 2HCl (aq)--> ZnCl2 (aq) + H2(g)

1 mole 2 moles

3.55gm of ore containing 37%Zn

mass of Zn = 3.55*37/100 =1.3135 gm

no of moles of Zn = 1.3135/65.4 = 0.02 moles

From the balanced equation

1 mole of Zn react with 2 moles of HCl

0.02 moles of Zn react with = 2*0.02/1 = 0.04 moles of HCl

Molarity = no of moles /volume in L

2.5 = 0.04/volume in L

volume in L = 0.04/2.5 = 0.016 L

volume in milliliters = 0.016*1000 = 16ml

Zinc reacts with hydrochloric acid according to the reaction
equation
Zn(s) + 2HCL (aq) -----------> ZnCl2(aq) + H2(g)
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solution in a coffee-cup calorimeter, all of the zinc reacts,
raising the temperature of the solution from 21.7 ∘C to 24.5 ∘C.
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the density of the solution and 4.18 J/g⋅∘C as the specific heat
capacity.)

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following balanced equation. Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) When
0.106 g of Zn(s) is combined with enough HCl to make 50.6 mL of
solution in a coffee-cup calorimeter, all of the zinc reacts,
raising the temperature of the solution from 21.5 ∘C to 24.4 ∘C.
Find ΔHrxn for this reaction as written. (Use 1.0 g/mL for the
density of the solution and 4.18 J/g⋅∘C as the specific heat
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following balanced equation. Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) When
0.106 g of Zn(s) is combined with enough HCl to make 54.5 mL of
solution in a coffee-cup calorimeter, all of the zinc reacts,
raising the temperature of the solution from 21.6 ∘C to 24.5 ∘C.
Find ΔHrxn for this reaction as written. (Use 1.0 g/mL for the
density of the solution and 4.18 J/g⋅∘C as the specific heat
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