1) What is the quanity og hydrogen gas liberated wheb 46 g sodium reacts with excess ethanol? (Given atomic mass of Na= 23) A) $2.4 \times 10^{-3} kg$ B) $2.0 \times 10^{-3} kg$ C) $4.0 \times 10^{-3} kg$ D) $2.4 \times 10^{-2} kg$ Answer: Option BExplanation: The reaction of ethanol with water is as follows $2C_{2}H_{5}OH(l)+2Na(s)\rightarrow 2C_{2}H_{5}\overline{O}Na^{+}+H_{2}(g)\uparrow$ 2 mole of Na (46g) =1 mole of H2 = 2g= 2 x 10-3 kg
