2Na + H2O == 2NaOH + H2...(1)
2Al + 2NaOH + 2H2O == 2NaAlO2 + 3H2...(2)
m(Al) = 0.27g n(Al) = 0.27/27 = 0.01mol
由反應(yīng)(2)生成的n2(H2) = 0.01/2 * 3 = 0.015mol
總共生成的n(H2) = 1.12/22.4 = 0.05mol
所以由反應(yīng)(1)生成的n1(H2) = 0.05 - 0.015 = 0.035mol
所以混合物中的n(Na) = 0.035 * 2 = 0.07mol
又因為溶液中還剩余n(NaOH) = 0.050 * 2 = 0.1mol
所以n(Na) + 2 * n(Na2O) = 0.01 + 0.1 = 0.11mol
所以n(Na2O) = 0.02mol
所以該鈉塊中鈉元素的質(zhì)量分數(shù) = (23 * 0.07 + 23 * 2 * 0.02)/(23 * 0.07 + 62 * 0.02) = 88.8%