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How many grams of zinc would you start with if you wanted to prepare 5.55 L of H2 at 288 mm Hg and 29.5 ∘C?

User Divick
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1 Answer

5 votes

Answer:

5.56g

Step-by-step explanation:

Given parameters:

Volume of H₂ = 5.55L

Pressure of H₂ = 288mmHg

Temperature of H₂ = 29.5°C

Unknown:

Mass of zinc = ?

Solution:

To start with, let us write the reaction equation of this simple displacement reaction:

Zn + 2HCl → ZnCl₂ + H₂

Now we have to work from the known or given compound to the unknown.

The known here is the hydrogen gas produced. From the parameters given, let us solve for the number of moles using the ideal gas law;

PV = nRT

take the given species to appropriate units;

288mmHg to atm gives;

760mmHg gives 1atm

288mmHg will give
(288)/(760) = 0.38atm

take the temperature to k;

29.5°C ; 29.5 + 273 = 302.5K

From the ideal gas equation:

P is the pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

Input the parameters and solve for n;

0.38 x 5.55 = n x 0.082 x 302.5

n = 0.085moles

From the balanced reaction equation;

1 mole of H₂ is produced from 1 mole of Zn

0.085moles of H₂ will be produced from 0.085 moles of Zn

Mass of Zn = number of moles of Zn x molar mass of Zn

= 0.085 x 65.4

= 5.56g

User Patrick Clay
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