Question

今晚上的作业,我又是插班生,很多知识还没学会,设xy）

Answer 1

表面被氧化的钠块含Na2O和钠.

4.8g水物质的量:4.8/18 = 0.27 mol

2Na + 2H2O = 2NaOH + H2(g)

2 2 2 1

x x x 2.24/22.4 = 0.1 mol

氢气的物质的量:2.24/22.4 = 0.1 mol

2 :1 = x :0.1

x = 0.2 mol

反应中消耗水和钠均为0.2 mol,生成0.2 mol NaOH (水为0.27mol,过量; 反应后剩水0.27-0.2 = 0.07 mol)

钠块中含钠:0.2* 23 = 4.6 g

钠块中含Na2O:5.22 - 4.6 = 0.62 g

Na2O物质的量:0.62/62 = 0.01 mol

Na2O + H2O = 2NaOH

1 1 2

0.01 y z

1 :1 = 0.01 :y

y = 0.01 mol (本反应所用水)

1 :2 = 0.01 :z

z = 0.02 mol (本反应生成NaOH)

溶液中溶质的物质的量(NaOH):0.2 + 0.02 = 0.22 mol

溶液中溶质的质量:0.22 * 40 = 8.8 g

溶液的质量:5.22 + 4.8 - 0.1*2 = 9.82 g

溶液中溶质的质量分数:8.8/9.82 = 89.6%