a)

$n_{Mg} = \dfrac{7,2}{24} = 0,3(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH : $n_{H_2} = n_{Mg} = 0,3(mol)$
$V_{H_2} = 0,3.24,79 = 7,437(lít)$

b)

$n_{MgCl_2} = n_{Mg} = 0,3(mol)$
$m_{MgCl_2} = 0,3.95 = 28,5(gam)$

c)

Gọi CTHH oxit là $R_2O_n$
$R_2O_n + nH_2 \xrightarrow{t^o} 2R + nH_2O$

Theo PTHH : $n_{oxit} = \dfrac{1}{n}.n_{H_2} = \dfrac{0,3}{n}(mol)$
$\Rightarrow \dfrac{0,3}{n}(2R + 16n) = 21,6$
$\Rightarrow R = 28n$

Với n = 2 thì R = 56(Fe)

Vậy oxit là $FeO$

nMg=7,224=0,3(mol)nMg=7,224=0,3(mol)
Mg+2HCl→MgCl2+H2Mg+2HCl→MgCl2+H2
nH2=nMg=0,3(mol)nH2=nMg=0,3(mol)
VH2=0,3.24,79=7,437(lít)VH2=0,3.24,79=7,437(lít)

b)

$n_{MgC{l}_2}=n_{Mg}=0,3\left(mol\right)$

a)\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

tỉ lệ       :1          2           1              1

số mol  :0,3       0,6        0,3           0,3

\(V_{H_2}=0,3.24,79=7,437\left(l\right)\)

b)\(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

c) Gọi CTHH của oxide là: \(R_xO_y\)

\(PTHH:R_xO_y+xH_2\xrightarrow[]{}xR+xH_2O\)

Theo phương trình ta có:\(2,48.\dfrac{1}{2}.\dfrac{R}{x}=21,6\)

\(\Rightarrow R_xO_y=80\left(đvc\right)\)

\(\Rightarrow R=56\Rightarrow Fe\)

Với x=2 và y=2 

thì CTHH có dạng \(FeO\)

help

Câu 5:

Ở 90 độ C, 100g   -     50g     -        150g dd

                   450g         225g    -        675g dd

Gọi số mol CuSO4.5H2O là x (mol)

\(\rightarrow m_{CuSO_4}=160x\left(g\right)\rightarrow m_{CuSO_4\left(còn\right)}=225-160x\left(g\right)\)

\(m_{H_2O}=180x\left(g\right)\rightarrow m_{H_2O\left(còn\right)}=450-180x\left(g\right)\)

Ở 10 độ C, 100 g nước thu được 36g CuSO4.

                 450-180x(g)             225-160x(g)

\(\rightarrow\left(450-180x\right).36=100.\left(225-160x\right)\)

\(\rightarrow16200-6480x=22500-16000x\)

\(16000x-6480x=22500-16200\)

\(\rightarrow x\approx0,662\)

\(m_{CuSO_4.5H_2O}=0,662.\left(160+180\right)=225\left(g\right)\)

b) m_H2SO4 ban đầu là bao nhiêu :) ?

\(S+O_2\underrightarrow{t^o}SO_2\)

\(1:1:1:1\)

\(0,2:0,2:0,2:0,2\left(mol\right)\)

\(n_{SO_2}=\dfrac{V}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

\(a,m_S=n.M=0,2.32=6,4\left(g\right)\)

\(b,V_{O_2}=n.24,79=0,2.24,79=4,958\left(l\right)\)

làm lại ko để ý có điều kiện=))))

\(n_{SO_2\left(dkc\right)}=\dfrac{V}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

\(PTHH:S+O_2-^{t^o}>SO_2\)

tỉ lệ        1   :     1     :     1

n(mol)  0,2<--0,2<---0,2

\(m_S=n\cdot M=0,2\cdot32=6,4\left(g\right)\\ V_{O_2\left(dkc\right)}=n\cdot24,79=0,2\cdot24,79=4,958\left(l\right)\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{0,54}{27}=0,02mol\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{22,05}{98}=0,225mol\)

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

0,02 <  0,225                                      ( mol )

0,02                                           0,03  ( mol )

\(V_{H_2}=n.22,4=0,03.24,79=0,7437l\)

\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

         0,05-->0,1------>0,05--->0,05

         FeO + H₂ --t^o--> Fe + H₂O

                  0,05------>0,05

=> \(\left\{{}\begin{matrix}m_{ZnCl_2}=0,05.136=6,8\left(g\right)\\V_{H_2}=0,05.24,79=1,2395\left(l\right)\\m_{Fe}=0,05.56=2,8\left(g\right)\end{matrix}\right.\)

\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,05                  0,05       0,05 
\(m_{ZnCl_2}=0,05.136=6,8\left(g\right)\\ V_{H_2}=0,05.24,79=1,2395l\) 
\(FeO+H_2\underrightarrow{t^o}Fe+H_2O\) 
 0,05   0,05   0,05 
\(m_{Fe}=0,05.56=2,8g\)

a, PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Ta có: \(n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,45\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,45.22,4=10,08\left(l\right)\)

b, Theo PT: \(n_{CO_2}=2n_{C_2H_4}=0,3\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,3.22,4=6,72\left(l\right)\)

c, PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{\downarrow}=m_{CaCO_3}=0,3.100=30\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\\ PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ n_{Al}=\dfrac{4}{2}.n_{Al_2O_3}=2.0,1=0,2\left(mol\right)\\ \Rightarrow x=m_{Al}=27.0,2=5,4\left(g\right)\\ b,n_{O_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,15.24,79=3,7185\left(l\right)\\ V_{kk\left(đktc\right)}=\dfrac{100}{20}.3,7185=18,5925\left(l\right)\)

a) \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

           0,2<---0,15<------0,1

=> X = 0,2.27 = 5,4 (g)

b) V_O2 = 0,15.24,79 = 3,7185 (l)

=> V_kk = 3,7185:20% = 18,5925(l)