Bài 4 : 

100ml = 0,1l

300ml = 0,3l

\(n_{Na2SO4}=0,2.0,1=0,02\left(mol\right)\)

\(n_{BaCl2}=0,1.0,3=0,03\left(mol\right)\)

Pt : \(Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4|\)

             1             1                2             1

         0,02          0,03           0,04        0,02           

a) Lập tỉ số so sánh : \(\dfrac{0,02}{1}< \dfrac{0,03}{1}\)

               ⇒ Na₂SO₄ phản ứng hết , BaCl₂ dư

               ⇒ Tính toán dựa vào số mol của Na₂SO₄

\(n_{BaSO4}=\dfrac{0,02.1}{1}=0,02\left(mol\right)\)

⇒ \(m_{BaSO4}=0,02.233=4,66\left(g\right)\)

b) \(n_{NaCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)

\(n_{BaCl2\left(dư\right)}=0,03-0,02=0,01\left(mol\right)\)

\(V_{ddspu}=0,1+0,3=0,4\left(l\right)\)

\(C_{M_{NaCl}}=\dfrac{0,04}{0,4}=0,1\left(M\right)\)

\(C_{M_{BaCl2\left(dư\right)}}=\dfrac{0,01}{0,4}=0,025\left(M\right)\)

 Chúc bạn học tốt

IV

1:

ĐKXĐ: \(x\in R\)

 \(3\sqrt{x^2-3x+5}+2\left(x+1\right)\left(x-4\right)=9\)

=>\(3\sqrt{x^2-3x+5}+2\left(x^2-3x-4\right)=9\)

=>\(3\sqrt{x^2-3x+5}+2\left(x^2-3x+5-9\right)=9\)

=>\(2\left(x^2-3x+5\right)+3\sqrt{x^2-3x+5}-27=0\)

=>\(2\left(x^2-3x+5\right)+9\sqrt{x^2-3x+5}-6\sqrt{x^2-3x+5}-27=0\)

=>\(\sqrt{x^2-3x+5}\left(2\sqrt{x^2-3x+5}+9\right)-3\left(2\sqrt{x^2-3x+5}+9\right)=0\)

=>\(\left(2\sqrt{x^2-3x+5}+9\right)\left(\sqrt{x^2-3x+5}-3\right)=0\)

=>\(\sqrt{x^2-3x+5}-3=0\)

=>\(\sqrt{x^2-3x+5}=3\)

=>\(x^2-3x+5=9\)

=>\(x^2-3x-4=0\)

=>(x-4)(x+1)=0

=>\(\left[{}\begin{matrix}x-4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

giải giúp bạn bài nào?

Quy hết kim loại R (II) và Fe về M (II):

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH: M + 2HCl ---> MCl₂ + H₂↑

           0,2<-0,4------->0,2---->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

\(b,M_M=\dfrac{8}{0,2}=40\left(\dfrac{g}{mol}\right)\\ \rightarrow M_R=2.40-56=24\left(\dfrac{g}{mol}\right)\)

=> R là Mg

c, Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

a---->2a--------->a

Mg + 2HCl ---> MgCl₂ + H₂

b---->2b-------->b

=> hệ pt \(\left\{{}\begin{matrix}56a+24b=8\\2a+2b=0,4\end{matrix}\right.\Leftrightarrow a=b=0,1\left(mol\right)\left(TM\right)\)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\end{matrix}\right.\)

d, \(\left\{{}\begin{matrix}C_{M\left(FeCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\\C_{M\left(MgCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\end{matrix}\right.\)

Gọi kim loại cần tìm là R (II)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_R=b\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow56a+M_R.b=8\left(g\right)\)

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH:

Fe + 2HCl --->FeCl₂ + H₂

a---->2a

R + 2HCl ---> RCl₂ + H₂

b---->2b

Theo pthh: \(n_{hhkl}=n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, Lập được hệ pt: \(\left\{{}\begin{matrix}56a+M_R.b=8\\2a+2b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\\M_R=24\left(\dfrac{g}{mol}\right)\end{matrix}\right.\)

=> R là Mg

c, \(\left\{{}\begin{matrix}n_{Fe}=0,1\left(mol\right)\\n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Mg}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

-14/18<0

0<-30/-40=3/4=9/12<9/11

9/11<1<-12/-8

=>-14/18<0<-30/-40<9/11<-12/-8