Bài 7:

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

b) Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{Zn\left(pư\right)}\)

\(\Rightarrow m_{Zn\left(pư\right)}=0,2\cdot65=13\left(g\right)\)

c) Theo PTHH: \(n_{HCl}=2n_{H_2}=0,4mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

(Coi như thể tích dd thay đổi không đáng kể)

PTHH: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

Gộp cả phần a và b

Ta có: \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)=n_{Ca\left(OH\right)_2}=n_{CaCO_3}\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\\m_{CaCO_3}=0,05\cdot100=5\left(g\right)\end{matrix}\right.\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)

 \(n_{H_2}=\dfrac{2,479}{22,4}=\dfrac{2479}{22400}mol\)

 \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo pt ta có: \(n_{Zn}=n_{H_2}=\dfrac{2479}{22400}mol\)\(\approx0,11mol\)

\(\Rightarrow m_{Zn}\approx7,2g\)

\(n_{HCl}=2n_{H_2}=0,22mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,22}{0,1}=2,2M\)

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) 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

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

a. PTHH: Fe + H₂SO₄ \(\rightarrow\) FeSO₄ + H₂

       TL:     1         1              1           1

      mol:  0,15  \(\leftarrow\) 0,15 \(\leftarrow\) 0,15  \(\leftarrow\) 0,15

\(b.m_{Fe}=n.M=0,15.56=8,4g\)

Đổi 150ml = 0,15 l

\(c.C_{MddH_2SO_4}=\dfrac{n}{V}=\dfrac{0,15}{0,15}=1M\)

a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Zn + 2HCl -----> ZnCl₂ + H₂

0,2     0,4               0,2     0,2

b, \(m_{Zn}=0,2.65=13\left(g\right)\)

\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

\(a,n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

         0,2<--0,4<--------0,2<---0,2

\(b,\left\{{}\begin{matrix}m_{Zn}=0,1.65=13\left(g\right)\\m_{HCl}=0,4.36,5=14,6\left(g\right)\end{matrix}\right.\\ c,m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)