Question

把表面已部分氧化为氧化钠的金属钠样品5.4g放入40g水中在标准状况下放出1.12l氢气

Answer 1

（1）产生氢气的物质的量= 1.12/22.4 = 0.05 mol

（2）原混合物中金属纳的质量为x克:

2Na + 2H2O = 2NaOH + H2

x = 23*n(Na) = 23*2*n(H2) = 23*2*0.05 = 2.3g

（3）Na2O 5.4-2.3 = 3.1g; n(Na2O) = 3.1/31 =0.1 mol

Na2O + H2O = 2NaOH

n(Na2O-NaOH)= 2n(Na2O) = 2*0.1 =0.2mol

n(Na-NaOH) = n(Na) = 2.3/23 =0.1mol

n(NaOH) = 0.2+0.1 = 0.3mol

若反应后溶液的体积为40ml,所得溶液的物质的量浓度=0.3/(40/1000) = 7.5 mol/L