$1) \\n_{CO_2}=\dfrac{8,96}{22,4}=0,4mol \\PTHH : \\Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2 \\CaCO_3+2HCl\to CaCl_2+H_2O+CO_2 \\Gọi\ n_{Na_2CO_3}=a;n_{CaCO_3}=b \\Ta\ có : \\m_{hh}=106a+100b=41,2g \\n_{CO_2}=a+b=0,4mol \\Ta\ có\ hpt : \\\left\{\begin{matrix} 106a+100b=41,2 & \\  a+b=0,4 &  \end{matrix}\right.⇔\left\{\begin{matrix} a=0,2 & \\  b=0,2 &  \end{matrix}\right. \\⇒\%m_{Na_2CO_3}=\dfrac{0,2.106}{41,2}.100\%=51,45\% \\\%m_{CaCO_3}=100\%-51,15\%=48,55\% \\b.Theo\ pt : \\n_{NaCl}=2.n_{Na_2CO_3}=2.0,2=0,4mol \\⇒m_{NaCl}=0,4.58,5=23,4g \\n_{CaCl_2}=n_{CaCO_3}=0,2mol \\⇒m_{CaCl_2}=0,2.111=22,2g \\∑n_{HCl}=2a+2b=2.0,2+2.0,2=0,8mol \\⇒m_{HCl}=0,8.36,5=29,2g \\⇒m_{dd\ HCl}=\dfrac{29,2}{20\%}=146g \\m_{dd\ spu}=41,2+146-0,4.44=169,6g \\⇒C\%_{NaCl}=\dfrac{23,4}{169,6}.100\%=13,8\% \\C\%_{CaCl_2}=\dfrac{22,2}{169,6}.100\%=13,1\% \\2) \\Na_2CO_3+Ba(OH)_2\to 2NaOH+BaCO_3↓$

Đáp án:

Giải thích các bước giải:

 1,  gọi nNa2co3=x

             ncaco3=y

    mhh=41,2 hay 106x+100y=41,2 (1)

      ta có ptpư naco3+2hcl-->2nacl+h20+co2

                        caco3+2hcl---->cacl2+h20+co2

theo pt nco2=nNaco3+ncaco3

=>  8,96/22,4=x+y

=>x+y=0,4 (2)

giải pt 1 , 2 ta đc : x=0,2 ,y=0,2(bn lm tiếp nha)

   2

a h2so4+ba(oh)2--->baso4+h20