把11.5g金属钠投入50g水中,反应为2Na+2H2O=2NaOH+H2,试计算如题 把11.5g金属钠投入50g水中,反应为2Na+2H2O=2NaOH+H2↑,试计算标准状况下下,①生成H2的体积是多少?②钠失去电子的总数约是多少个?③所得溶液中溶质的质量分数是多少?④若所得溶液的密度为1.1g/cm3,溶液中溶质的物质的量浓度是多少?

n(Na)= 11.5 g / 23g/mol = 0.5 mol n(H2O) = 50g / 18g/mol = 2.78mol 所以水过量 (1)n(H2) = 1/2 n(Na) = 1/2 X 0.5mol = 0.25 mol V(H2) = 0.25mol X 22.4L/mol = 5.6L (2)n(e-) = n(Na) = 0.5mol (3)n(NaOH) = n(Na) = 0.5mol m(NaOH) = 0.5mol X 40g/mol = 20g m(H2) = 0.25 mol X 2g/mol = 0.5g m(溶液) = 11.5g + 50 g - 0.5g = 61g w = 20 g / 61g X 100% = 32.8% (4) V(溶液) = 61g / 1.1g/cm3 = 55.45mL = 0.0555L c(NaOH) = 0.5 mol / 0.0555L = 9.02 mol/L