1) What is the quanity og hydrogen gas liberated wheb 46 g sodium reacts with excess ethanol? (Given atomic mass of Na= 23) A) 2.4×10−3kg B) 2.0×10−3kg C) 4.0×10−3kg D) 2.4×10−2kg Answer: Option BExplanation: The reaction of ethanol with water is as follows 2C2H5OH(l)+2Na(s)→2C2H5¯ONa++H2(g)↑ 2 mole of Na (46g) =1 mole of H2 = 2g= 2 x 10-3 kg
