Question

Aluminum reacts with excess aqueous hydrochloric acid to produce hydrogen. (a) Write a balanced chemical equation for the reaction. (Hint: Water-soluble AlCl3 is the stable chloride of aluminum.) (b) Calculate the mass of pure aluminum that will furnish 10.0 L hydrogen at a pressure of 0.750 atm and a temperature of 30.0°C.

Answer 1

Answer:

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

5.39 g

Explanation:

The balanced equation for the reaction of aluminum with hydrochloric acid is shown below as:-

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

Given,

Pressure = 0.750 atm

Temperature = 30.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (30.0 + 273.15) K = 303.15 K  

T = 303.15 K  

Volume = 10.0 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.750 atm × 10.0 L = n × 0.0821 L.atm/K.mol × 303.15 K  

⇒n = 0.3013 moles

Moles of hydrogen obtained = 0.3013 moles

From the reaction,

3 moles of hydrogen gas are furnished when 2 moles of aluminum is consumed.

Also,

1 mole of hydrogen gas are furnished when $\frac{2}{3}$ mole of aluminum is consumed.

Thus,

0.3013 mole of hydrogen gas are furnished when $\frac{2}{3}\times 0.3013$ mole of aluminum is consumed.

Moles of aluminum consumed = 0.2 moles

Also, Molar mass of aluminum = 26.981539 g/mol

So, Mass = Moles*Molar mass = 0.2 moles*26.981539 g/mol  =  5.39 g