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Please help with 2nd and 3rd questions In question b) Aluminum reacts with hydrocloric acid

Please help with 2nd and 3rd questions In question b) Aluminum reacts with hydrocloric-example-1
User Jaycee
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1 Answer

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In order to solve the question b):

We have the following reaction occuring:


2Al+6HCl\rightarrow2AlCl_3+3H_2

We need to calculate the number of moles of aluminum needed to produce 30ml of H2 (hydrogen gas) at a temperature of 22°C and a pressure of 763mmHg.

To calculate the number of moles of hydrogen produced we use the ideal gas equation:


P.V=n.R.T

Where:

P is the pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

R is a constant and it's value is:


R=8.314\text{ }(m^3.Pa)/(K.mol)

So we need to convert each variable to the units of this constant.


\begin{gathered} T:\text{ }22^(\circ)C=295.15^(\circ)K \\ P:\text{ }763mmHg=101725\text{ }Pa \\ V:\text{ 30ml}:0.00003m^3 \end{gathered}

So now we calculate:


n=(P.V)/(R.T)=(101725Pa.0.00003m^3)/(8.314(m^3.Pa)/(K.mol).295.15^(\circ)C)=0.00124mol

So we know that 0.00124 moles of H2 are formed.

Now we know that for every 2 moles of aluminum 3 moles of H2 are formed.

So we calculate the moles of Al needed:


n_(Al)=0.00124mol_(H2).(2mol_(Al))/(3mol_(H2))=0.000826mol_(Al)

So the answer is 0.000826 moles of Al°

User Mike Vella
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