CaBr2 → CaCl2 dm
200 160-71
x 13.4-11.175
x==200*(13.4-11.175)/(160-71)== 5
即 混合物中 CaBr2为 5 g ,则 CaCl2为 13.4-5==8.4 g
n(CaBr2)==5/200==0.025 mol n(CaCL2)==8.4/111=0.0757
Ca2+ :Cl- :Br- ==0.1 ::0.15 :0.05==2:3:1
选 D
设要加入 x ml
则H2SO4 总物质的量===1.84*x*98%/98 +0.5*0.2
K2SO4物质的量不变==0.5*0.2=0.1
则 (1.84*x*98%/98 +0.5*0.2) :0.1 ==2 :0.1
x==103.3 ml