solution:
Aluminum responds with scathing soft drink as per the response
The response of aluminum with scathing soft drink can be addressed as:
\[2Al\text{ }+\text{ }2NaOH\text{ }+\text{ }2H2O\text{ }\to \text{ }2NaAlO2\text{ }+\text{ }3H2\]
Thusly volume of hydrogen at STP delivered when 0.15g of Al reactsv
\[\begin{array}{*{35}{l}}
=0.15\text{ }x\text{ }3\text{ }x\text{ }22.4/54\text{ }=\text{ }187ml \\
~ \\
Presently\text{ }P1~=\text{ }1\text{ }bar, \\
~ \\
P2~=\text{ }1\text{ }bar \\
~ \\
T1~=\text{ }273\text{ }K \\
~ \\
T2~=\text{ }20\text{ }+\text{ }273\text{ }=\text{ }293\text{ }K \\
~ \\
V1~=\text{ }187\text{ }ml \\
~ \\
V2~=\text{ }x \\
\end{array}\]
At the point when strain is held constant,then
\[V2~=\text{ }P1~V1~T2~/P2~T1\]
Or then again
\[x\text{ }=\text{ }1\text{ }X\text{ }187\text{ }X\text{ }293~/~0.987\text{ }X\text{ }273\text{ }=\text{ }201\text{ }ml\]
Thusly, 201 mL of dihydrogen will be delivered