a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

            0,1--->0,1------->0,1------>0,1

=> \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b) \(m_{\text{dd}H_2SO_4}=\dfrac{0,1.98}{19,6\%}=50\left(g\right)\)

=> \(m_{\text{dd}.sau.p\text{ư}}=50+6,5-0,1.2=56,3\left(g\right)\)

=> \(C\%_{ZnSO_4}=\dfrac{0,1.161}{56,3}.100\%=28,6\%\)

a)\(n_{MgO}\)=6:40=0,15(mol)

Ta có PTHH:

MgO+\(H_2SO_4\)->MgS\(O_4\)+\(H_2O\)

0,15......0,15...........0,15..................(mol)

Theo PTHH:\(m_{H_2SO_4}\)=0,15.98=14,7g

b)Ta có:\(m_{ddH_2SO_4}\)=D.V=1,2.50=60(g)

=>Nồng độ % dd \(H_2SO_4\) là:

\(C_{\%ddH_2SO_4}\)=\(\dfrac{14,7}{60}\).100%=24,5%

c)Theo PTHH:\(m_{MgSO_4}\)=0,15.120=18(g)

Khối lượng dd sau pư là:

\(m_{ddsau}\)=\(m_{MgO}\)+\(m_{ddH_2SO_4}\)=6+60=66(g)

Vậy nồng độ % dd sau pư là:

\(C_{\%ddsau}\)=\(\dfrac{18}{66}\).100%=27,27%

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

        1           1              1            1

      0,15       0,15                        0,15

\(n_{H2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

\(n_{H2SO4}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

300ml = 0,3l

\(C_{MddH2SO4}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)

 Chúc bạn học tốt

n_H2SO4 = 49/98 = 0.5 (mol) 

C_MH2SO4 = 0.5/0.15 = 3.3 (M) 

Zn + H₂SO₄ => ZnSO₄  + H₂

...........0.5.............0.5.........0.5

V_H2 = 0.5 * 22.4 = 11.2 (l) 

C_MZnSO4 = 0.5 / 0.15 = 10/3 (M) 

C%_ZnSO4 = C_M*M / 10D = 10/3 * 161 / 10 * 1.25 = 42.9 %

Anh bổ sung câu c)

\(C_{MddNa_2SO_4}=\dfrac{0,25}{0,09879+0,5}=0,4175\left(M\right)\)

Bổ sung: \(D_{HCl}=1,18\left(g/ml\right)\)

a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\\n_{HCl}=\dfrac{100\cdot1,18\cdot20\%}{36,5}=\dfrac{236}{365}\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{\dfrac{236}{365}}{2}\) \(\Rightarrow\) HCl còn dư, MgO p/ứ hết

\(\Rightarrow n_{MgCl_2}=0,05\left(mol\right)\) \(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)