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sergejj [24]
2 years ago
13

A laboratory assistant needs to prepare 35.2 liters of hydrogen at 25.0°c and 101.3 kilopascals. this is the equation for the re

action: 2hcl ca → h2 cacl2 what volume of 2.3 m hydrochloric acid is required to produce this much gas? use the ideal gas resource.
Chemistry
1 answer:
Sedaia [141]2 years ago
5 0

Volume of hydrogen gas = 35.2L

Temperature of hydrogen gas = 25.0°C + 273 = 298 K

Pressure of hydrogen gas = 101.3 kPa = 1 atm

(Conversion factor 1 atm = 101.325 kPa)

Using the Ideal gas law,

PV = nRT

where P, V, n, R and T are pressure, volume, number of moles, universal gas constant and temperature respectively.

n = PV/RT

The number of moles of hydrogen gas,

n = 35.2 L x 1 atm/ 0.0821 L.atm.mol⁻¹1.K⁻¹ x 298 K

n = 1.438 moles

2HCl + Ca → H₂ + CaCl₂

Based on the given balanced equation, 1 mole of H₂ is produced from two  moles of HCl.

Since H₂ is 1.438 moles, HCl will be (1.438 x 2) moles or 2.876 moles

Given, Molarity of HCl = 2.3 M

Molarity = moles of solute / liter of the solution

Thus the volume of HCl is,

2.3 M = 2.876 moles / liter of the solution

V = 2.876 moles / 2.3 M

V = 1.25 L

Thus the volume of 2.3 M hydrochloric acid required to produce the given amount gas is 1.25 L.

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The two half reactions are:

The oxidation reaction: Zn(s) → Zn²⁺(aq) + 2e.

The reduction reaction: Sn²⁺(aq) + 2e → Sn(s).

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So, the net chemical equation is:

<em>Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).</em>

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The two half reactions are:

The oxidation reaction: Mg(s) → Mg²⁺(aq) + 2e.

The reduction reaction: Cr³⁺(aq) + 3e → Cr(s).

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<em>3Mg(s) + 2Cr³⁺(aq) → 3Mg²⁺(aq) + 2Cr(s).</em>

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The two half reactions are:

The oxidation reaction: Al → Al³⁺ + 3e.

The reduction reaction: MnO₄⁻ + 8H⁺ + 5e → Mn²⁺ + 4H₂O.

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