The reaction of aluminum with caustic soda is as given below:

2Al + 2NaOH + 2H_{2}O → 2NaAlO_{2} + 3H_{2}

At Standard Temperature Pressure

(273.15 K and 1 atm), 54 g ( 2 × 27 g)

of Al gives 3 ×22400 mL of H_{2}

Therefore, 0.15 g Al gives:

= \( \dfrac{3 \; \times \; 22400 \; \times \; 0.15}{54}mL \) of H_{2}

= 186.67 mL of H_{2}

At Standard Temperature Pressure,

\( p_{ 1 }\) = 1 atm

\( V_{ 1 }\) = 186.67 mL

\( T_{ 1 }\)= 273.15 K

Let the volume of dihydrogen be

\( V_{ 2 } \;at\; p_{ 2 }\)

= \( 0.987 \;atm (since\; 1\; bar = 0.987 atm) \)

\( and\; T_{ 2 }= 20^{\circ} C\)

\( = (273.15 + 20) K \)

\( = 293.15 K.\)

Now,

\( \dfrac{p_{ 1 }V_{ 1 }}{T_{ 1 }}= \dfrac{p_{ 2 }V_{ 2 }}{T_{ 2 }}\)

\( V_{ 2 } = \dfrac{p_{ 1 }V_{ 1 }T_{ 2 }}{p_{ 2 }T_{ 1 }}\)

=\( \dfrac{1 \; \times \; 186.67 \; \times \; 293.15}{0.987 \; \times \; 273.15}\)

= 202.98 mL

= 203 mL

Hence, 203 mL of dihydrogen will be released.

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