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m\(_{H_2SO_{4\left(sau\right)}}=\dfrac{12.200}{100}+\dfrac{300.40}{100}=144\left(g\right)\)
\(M_{dd\left(sau\right)}=200+300=700\left(g\right)\)
\(\rightarrow C\%_{\left(sau\right)}=\dfrac{124}{500}.100=28,8\%\)
\(\Rightarrow\)chọn B
Với 200g dd H2SO4 9.8% ==> mH2SO4 = 200x9.8/100 = 19.6 (g)
Với 300g dd H2SO4 19.6% => mH2SO4 = 300x19.6/100 = 58.8 (g)
mdd H2SO4 = 200 + 300 = 500 (g)
C% X = (19.6+58.8)x100/500 = 15.68%
mH2SO4 = 19.6 + 58.8 = 78.4 (g)
===> nH2SO4 = m/M = 78.4/98 = 0.8 (mol)
H2SO4 + 2KOH => K2SO4 + 2H2O
==> nKOH = 0.8 x 2 = 1.6 (mol)
V = n/CM = 1.6/1 = 1.6 (l) = 1600ml
Ta có
\(m_{BaCl_2}=\frac{200.20,8}{100}=41,6\left(g\right)\Rightarrow n_{BaCl_2}=\frac{41,6}{208}=0,2\left(mol\right)\)
\(m_{Na_2SO_4}=\frac{150.28,4}{100}=42,6\left(g\right)\Rightarrow n_{Na_2SO_4}=\frac{42,6}{142}=0,3\left(mol\right)\)
\(BaCl_2+Na_2SO_4\rightarrow BaSO_4\downarrow+2NaCl\)
Thấy \(n_{BaCl_2}< n_{Na_2SO_4}\Rightarrow\) Tính theo \(BaCl_2\)
\(m_{ddsaupu}=200+150=350\left(g\right)\)
Có \(n_{NaCl}=2n_{BaCl_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaCl}=58,5.0,4=23,4\left(g\right)\Rightarrow C\%\left(NaCl\right)=\frac{23,4}{350}.100\%\approx6,67\%\)
BaCl2 + Na2SO4-----> BaSO4 +2 NaCl
Ta có
m\(_{BaCl2}=\)\(\frac{200.20,8}{100}=41,6\left(g\right)\)
n\(_{BaCl2}=\frac{41,6}{110}=0,38\left(mol\right)\)
m\(_{Na2SO4}=\frac{150.28,4}{100}=42,6\left(g\right)\)
n\(_{Na2SO4}=\frac{42,6}{142}=0,3\left(mol\right)\)
=> BaCl2 dư
Theo pthh
n\(_{BaCl2}=n_{Na2SO4}=0,3\left(Mol\right)\)
n\(_{BaCl2}dư=0,38-0,3=0,08\left(mol\right)\)
mdd= 200+150=350(g)
C%(BaCl2)=\(\frac{0,08.110}{350}.100\%=2,5\%\)
Theo pthh
n\(_{NaCl}=2n_{Na2SO4}=0,6\left(mol\right)\)
C%(NaCl)=\(\frac{0,6.58,5}{350}.100\%=10\%\)
Chúc bạn học tốt
Ta có:
\(n_{HCl\left(D\right)}=n_{NaOH}=\frac{0,1.15}{1000}=0,0015\left(mol\right)\)
\(PTHH:NaOH+HCl\rightarrow NaCl+H_2O\)
\(AgNO_3+HCl\rightarrow AgCl+HNO_3\)
\(\Rightarrow CM_D=\frac{0,0015}{0,01}=0,15M\)
\(n_{HCl\left(E\right)}=n_{AgCl}=\frac{2,87}{143,5}=0,02\left(mol\right)\)
\(\Rightarrow CM_E=\frac{0,02}{0,08}=0,25M\)
Giải hệ PT:
\(\left\{{}\begin{matrix}3a+b=10\\a+3b=0,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}CM_A=0,3M\\CM_B=0,1M\end{matrix}\right.\)
\(2A+H_2SO_4\rightarrow A_2SO_4+H_2\\ Ta.có:n_A=2n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow M_A=\dfrac{7,8}{0,2}=39\left(K\right)\\ n_{H_2SO_4}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow CM_{H_2SO_4}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
\(a.2A+H_2SO_4\rightarrow A_2SO_4+H_2\\ Tacó:n_A=2n_{H_2}=0,2\left(mol\right)\\ \Rightarrow M_A=\dfrac{7,8}{0,2}=39\left(K\right)\\ n_{H_2SO_4}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow CM_{H_2SO_4}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
Đáp án C