a, Ta có: \(C\%_{ddH_2SO_4}=\dfrac{75}{100}.100\%=75\%\)

b, Ta có: \(C_{M_{ddHCl}}=\dfrac{0,2}{2}=0,1M\)

Bạn tham khảo nhé!

\(n_{OH^-}=n_{NaOH}=0,3.1,5=0,45\left(mol\right)\\ n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,2x+0,5.0,2.2=0,2x+0,2\left(mol\right)\)

PT ion rút gọn: \(H^++OH^-\rightarrow H_2O\)

                      0,45<---0,45

\(\Rightarrow0,2x+0,2=0,45\Leftrightarrow x=1,25M\)

Ta có: \(V_{dd}=0,3+0,2=0,5\left(l\right)\) và \(\left\{{}\begin{matrix}n_{Na^+}=0,45\left(mol\right)\\n_{Cl^-}=0,2.1,25=0,25\left(mol\right)\\n_{SO_4^{2-}}=0,2.0,5=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{Na^+}=\dfrac{0,45}{0,5}=0,9M\\C_{Cl^-}=\dfrac{0,25}{0,5}=0,5M\\C_{SO_4^{2-}}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

Bạn thử kiểm tra lại xem đề có thiếu không nhé vì bạn chưa cho C_M của dung dịch NaOH hoặc thể tích H2SO4

\(n_{SO_3}=\dfrac{30}{80}=0,375\left(mol\right)\)

PTHH: SO₃ + H₂O ---> H₂SO₄

        0,375 ----------------> 0,375

\(C_{M\left(H_2SO_4\right)}=\dfrac{0,375}{0,15}=2,5M\)

PTHH: 2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

           0,25 <- 0,375

\(m_{Al}=0,25.27=6,75\left(g\right)\)

\(\left\{{}\begin{matrix}n_{NaOH}=0,3.2=0,6\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,2.0,5=0,1\left(mol\right)\\n_{H_2SO_4}=0,2.0,5=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}n_{Na^+}=0,6\left(mol\right)\\n_{Fe^{3+}}=0,1.2=0,2\left(mol\right)\\n_{H^+}=0,1.2=0,2\left(mol\right)\end{matrix}\right.\\\left\{{}\begin{matrix}n_{SO_4^{2-}}=0,1.3+0,1=0,4\left(mol\right)\\n_{OH^-}=0,6\left(mol\right)\end{matrix}\right.\end{matrix}\right.\)

PT ion rút gọn:

\(H^++OH^-\rightarrow H_2O\)

0,2-->0,2

\(Fe^{3+}+3OH^-\rightarrow Fe\left(OH\right)_3\downarrow\)

\(\dfrac{2}{15}\)<----0,4--------->\(\dfrac{2}{15}\)

\(\Rightarrow m=\dfrac{2}{15}.107=\dfrac{214}{15}\left(g\right)\)

dd sau phản ứng có: \(\left\{{}\begin{matrix}n_{Na^+}=0,6\left(mol\right)\\n_{Fe^{3+}\left(d\text{ư}\right)}=0,2-\dfrac{2}{15}=\dfrac{1}{15}\left(mol\right)\\n_{SO_4^{2-}}=0,4\left(mol\right)\end{matrix}\right.\)

\(V_{\text{dd}}=0,3+0,2=0,5\left(l\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{Na^+}=\dfrac{0,6}{0,5}=1,2M\\C_{Fe^{3+}}=\dfrac{\dfrac{1}{15}}{0,5}=\dfrac{2}{15}M\\C_{SO_4^{2-}}=\dfrac{0,4}{0,5}=0,8M\end{matrix}\right.\)

Fe + H₂SO₄ → FeSO₄ + H₂ (1)

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂ (2)

\(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

Gọi x,y lần lượt là số mol của Fe và Al

Ta có: \(56x+27y=11\) (*)

Theo PT1: \(n_{H_2}=n_{Fe}=x\left(mol\right)\)

Theo Pt2: \(n_{H_2}=\frac{3}{2}n_{Al}=1,5y\left(mol\right)\)

Ta có: \(x+1,5y=0,4\) (**)

Từ (*)(**) ta có: \(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

Vậy \(n_{Fe}=0,1\left(mol\right)\)

\(n_{Al}=0,2\left(mol\right)\)

Theo Pt1: \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)

Theo pt2: \(n_{H_2SO_4}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0,2=0,3\left(mol\right)\)

\(\Rightarrow\Sigma n_{H_2SO_4}=0,1+0,3=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\frac{0,4}{0,2}=2\left(M\right)\)

Gọi số mol Fe và Al là x và y

\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

(mol) 1 1 1 1

(mol) x x x x

\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

(mol) 2 3 1 3

(mol) y 3y/2 y 3y/2

Ta có hpt:

\(\left\{{}\begin{matrix}56x+27y=11\\22,4x+33,6y=8,96\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Fe}=0,1\left(mol\right)\\n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)