设混合物中碳酸氢钠为xg,则
2NaHCO3
Na2CO3+CO2↑+H2O△m
△
168 106 62
x (3.80g-3.18g)
=168 x
,解得x=1.68g,62 3.80g?3.18g
n(NaHCO3)=
=0.01mol,1.68g 168g/mol
m(Na2CO3)=3.80g-1.68g=2.12g,
n(Na2CO3)=
=0.02mol,2.12g 106g/mol
所以混合物中碳酸钠和碳酸氢钠的物质的量之比为0.02mol:0.01mol=2:1,
故选B.