Question

Use the equation Zn + 2HCl -> ZnCl_{2} +
H_{2}

_{If you react 71.24 g Zn with 91.53 g HCl, what mass of zinc
chloride is produced? And what is the limiting reactant in this
situation? Show all calculations.}

Answer #1

Zn + 2HCl -> ZnCl2 + H2

65.3 g 2 x 36.5 g = 73 g

Given that 71.24 g Zn reacted with 91.53 g HCl.

Hence,

Moles of Zn = mass / molar mass = 71.24 g / 65.3 g/mol = 1.09 mol

Moles of HCl = mass / molar mass = 91.53 g / 73 g/mol = 1.25 mol

Here, Zn present in lesser no of moles. Therefore, Zn is the limiting reagent and yield is calculated based on Zn.

Calculation of mass of zinc chloride produced

Zn + 2HCl -> ZnCl2 + H2

65.3 g 136.3 g

71.24 g ?

? = (71.24 g Zn/ 65.3 g Zn ) x 136.3 g ZnCl2

= 148.7 g ZnCl2

mass of ZnCl2 = 148.7 g ZnCl2

Therefore, 148.7 g ZnCl2 wil be produced.

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?

Zinc reacts with hydrochloric acid according to the reaction
equation
Zn(s) + 2HCL (aq) -----------> ZnCl2(aq) + H2(g)
How many milliliters of 2.00 M HCl(aq) are required to react
with 7.05 g of an ore containing 38.0% Zn(s) by mass?

Zinc reacts with hydrochloric acid according to the reaction
equation
Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)
How many milliliters of 6.50 M HCl(aq) are required to react
with 3.05 g of Zn(s)?

Zinc metal reacts with hydrochloric acid according to the
following balanced equation. Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) When
0.103 g of Zn(s) is combined with enough HCl to make 55.0 mL of
solution in a coffee-cup calorimeter, all of the zinc reacts,
raising the temperature of the solution from 21.7 ∘C to 24.4
∘C.

Zinc metal reacts with hydrochloric acid according to the
following balanced equation. Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) When
0.107 g of Zn(s) is combined with enough HCl to make 51.6 mL of
solution in a coffee-cup calorimeter, all of the zinc reacts,
raising the temperature of the solution from 22.2 ∘C to 24.3 ∘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

Zinc metal reacts with hydrochloric acid according to the
following balanced equation. Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) When
0.119 g of Zn(s) is combined with enough HCl to make 53.4 mL of
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.
Part A 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
capacity.)

Zinc metal reacts with hydrochloric acid according to the
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
capacity.)

A 2.25 gram sample of zinc metal reacts with 5.00 grams of
hydrochloric acid to give zinc chloride and hydrogen gas according
the blanced equation:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Molar mass ZnCl2: 136.28
g/mol
a. What is the limiting reactant?
b. What mass of ZnCl2 can be formed? (This is called the
theoretical yield.)

Zn + HCl --> 2ZnCl2 + H2
Assuming you start with 10.0 g of zinc and 10.0 g hydrochloric
acid, identify the limiting reagent and determine what mass of the
other reagent will remain after the reaction runs to
completion.

What volume (in mL!!!) of 2.25 M HCl is required to react with
2.03 g of zinc (65.41 g/mol) according to the following reaction?
Zn(s) + 2 HCl (aq) → ZnCl2(aq) + H2(g)

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