a) PTHH: Zn + H2SO4 -> ZnSO4 + H2

nH2= 0,15(mol)

=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)

b) mZn=0,15.65=9,75(g)

c) CMddH2SO4= 0,15/ 0,05=3(M)

d) mZnSO4= 161. 0,15=24,15(g)

zn+ h2so4-> znso4+ h2

nh2=3,36/22,4=0,15

nzn= nh2=0,15mol

-> mzn=0,15*65=9,75g

nh2so4=nh2=0,15

cM h2so4=0,15/0,05=3M

nznso4=nh2=0,15

mznso4=0,15*161=24,15g

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo ĐLBTKL:

m_Zn + m_HCl = m_H2 + m_ZnCl2

=> m_HCl = 40,8 + 0,3.2 - 19,5 = 21,9(g)

\(Zn+2HCl->ZnCl_2+H_2\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

 Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)

b, m_Zn = 0,15.65 = 9,75 (g)

c, C_{M (H2SO4)} = 0,15/0,05 = 3 M

d, m_ZnSO4 = 0,15.161 = 24,15 (g)

Bạn tham khảo nhé!

a, \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: Cu + H₂SO₄ l → ko phản ứng

PTHH: Fe + H₂SO₄ → FeSO₄ + H₂

Mol:   0,05      0,05                      0,05

\(m_{Fe}=0,05.56=2,8\left(g\right)\Rightarrow m_{Cu}=3,44-2,8=0,64\left(g\right)\)

b, \(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)    

giúp mình giải với

400ml = 0,4l

\(n_{H2SO4}=0,5.0,4=0,2\left(mol\right)\)

Pt : \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2|\)

        1           1               1            1

     0,2                           0,2

\(n_{ZnSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnSO4}=0,2.161=32,2\left(g\right)\)

 Chúc bạn học tốt

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

Ta có: \(n_{H_2SO_4}=0,2\cdot1,5=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2\cdot27=5,4\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

            \(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)

Ta có: \(n_{SO_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)=n_{Cu}\)

\(\Rightarrow m_{Cu}=0,005\cdot64=0,32\left(g\right)\) \(\Rightarrow m_{Mg}=10,6-0,32=10,28\left(g\right)\)

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

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g

a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)

\(m_{Al}=0,1.27=2,7\left(g\right)\)

c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

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

0,1----------------------0,075----0,15

n H2=0,15 mol

=>mAl=0,1.27=2,7g

=>m Al2(SO4)3=0,075.342=25,65g