$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CO_2} = n_{CaCO_3} = \dfrac{15}{100} = 0,15(mol)$
$n_{NaOH} = \dfrac{80.12,5\%}{40} = 0,25(mol)$

Ta thấy \(1< \dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,25}{0,15}=1,67< 2\) nên X gồm $Na_2CO_3(a\ mol);NaHCO_3( b\ mol)$

$2NaOH + CO_2 \to Na_2CO_3 + H_2O$
$NaOH + CO_2 \to NaHCO_3$

Ta có : $2a + b = 0,25$ và $a + b = 0,15$
Suy ra : a = 0,1 ; b = 0,05

$m_{dd\ X} = m_{CO_2} + m_{dd\ NaOH} = 86,6(gam)$
$C\%_{Na_2CO_3} = \dfrac{0,1.106}{86,6}.100\% = 12,24\%$

$C\%_{NaHCO_3} = \dfrac{0,05.84}{86,6}.100\% = 4,85\%$

Giải bài này giúp mk ik!!!!!!

a) 

Mg + 2HCl ---> MgCl₂ + H₂ (1)

MgO + 2HCl ---> MgCl₂ + H₂O (2)

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

Theo (1): \(n_{Mg}=n_{H_2}=0,15\left(mol\right)\)

=> \(\left\{{}\begin{matrix}m_{Mg}=0,15.24=3,6\left(g\right)\\m_{MgO}=10-3,6=6,4\left(g\right)\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{3,6}{10}.100\%=36\%\\\%m_{MgO}=100\%-36\%=64\%\end{matrix}\right.\)

d) \(n_{MgO}=\dfrac{6,4}{40}=0,16\left(mol\right)\)

Theo (1), (2): \(n_{HCl}=2n_{Mg}+2n_{MgO}=0,62\left(mol\right)\)

=> \(C_{M\left(HCl\right)}=\dfrac{0,62}{0,2}=3,1M\)

e) Theo (1), (2): \(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}=0,31\left(mol\right)\)

=> m_muối = 0,31.95 = 29,45 (g)

f) \(C_{M\left(MgCl_2\right)}=\dfrac{0,31}{0,2}=1,55M\)

Lớp 1 có Hóa Học ư

không ngờ lớp 1 cũng có hóa học đấy ;))

1) 

$Fe + 2HCl \to FeCl_2 + H_2$
$CuO + 2HCl \to CuCl_2 + H_2O$

Theo PTHH, $n_{Fe} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$

$\%m_{Fe} = \dfrac{0,25.56}{30}.100\% = 46,67\%$

$\%m_{CuO} = 100\% - 46,67\% = 53,33\%$

2)

$n_{CuO} = \dfrac{30 - 0,25.56}{80} = 0,2(mol)$

Ta có : $n_{HCl} =2n_{Fe} + 2n_{CuO} = 0,9(mol)$
$\Rightarrow V_{dd\ HCl} =\dfrac{0,9}{1,6} = 0,5625(lít)$

1) \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂ 

          0,25<-0,5<------0,25<----0,25

=> \(\left\{{}\begin{matrix}m_{Fe}=0,25.56=14\left(g\right)\\n_{CuO}=30-14=16\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{14}{30}.100\%=46,67\%\\\%m_{CuO}=100\%-46,67\%=53,33\%\end{matrix}\right.\)

2) \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

PTHH: CuO + 2HCl ---> CuCl₂ + H₂O

            0,2----->0,4

=> \(V_{ddHCl}=\dfrac{0,4+0,5}{1,6}=0,5625M\)

1) 

$Fe + 2HCl \to FeCl_2 + H_2$
$CuO + 2HCl \to CuCl_2 + H_2O$

Theo PTHH, $n_{Fe} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$

$\%m_{Fe} = \dfrac{0,25.56}{30}.100\% = 46,67\%$

$\%m_{CuO} = 100\% - 46,67\% = 53,33\%$

2)

$n_{CuO} = \dfrac{30 - 0,25.56}{80} = 0,2(mol)$

Ta có : $n_{HCl} =2n_{Fe} + 2n_{CuO} = 0,9(mol)$
$\Rightarrow V_{dd\ HCl} =\dfrac{0,9}{1,6} = 0,5625(lít)$

Gọi n_Fe = x ; n_CuO = y (mol) 

Phương trình : Fe + 2HCl --> FeCl₂ + H₂ 

                         x    -> 2x                   -> x 

CuO + 2HCl ---> CuCl₂ + H₂O                   

Lại có \(x=n_{H_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\)(mol) 

=> \(m_{Fe}=n.M=0,25.56=14\left(g\right)\)

=> \(m_{Cu}=30-14=16\left(g\right)\)

=> \(\%Fe=46,7\%;\%Cu=53,3\%\)

b) \(V_{Hcl}=\dfrac{n}{C_M}=\dfrac{0,5}{1,6}=\dfrac{5}{16}\left(l\right)\)

a. \(n_{H_2}=\dfrac{6.72}{22,4}=0,3\left(mol\right)\)

PTHH : Mg + 2HCl -> MgCl₂ + H₂

PTHH : Fe + 2HCl -> FeCl₂ + H₂

Gọi \(n_{Mg}=a\left(mol\right);n_{Fe}=b\left(mol\right)\)

\(\Rightarrow24a+56b=10,4\left(g\right)\left(1\right)\)

\(\Rightarrow a+b=0,3\left(mol\right)\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrow a=0,2\left(mol\right),b=0,1\left(mol\right)\)

\(\%m_{Mg}=\dfrac{0,2.24}{10,4}=46,1\%\)

\(\%m_{Fe}=100\%-46,1\%=53,9\%\)

b. \(n_{hh}=0,1+0,2=0,3\left(mol\right)\)

Theo PT : \(n_{HCl}=2n_{hh}=0,6\left(mol\right)\)

\(C\%_{HCl}=\dfrac{0,6.36,5}{300}.100=7,3\%\)

c. Theo PT : \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)

                     \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{10,4+300-0,6}=6,13\%\\C\%_{FeCl_2}=\dfrac{0,1.127}{10,4+300-0,6}=4,09\%\end{matrix}\right.\)

Ta có : $2p + n = 13 \Rightarrow n = 13 - 2p$

$1 ≤ \dfrac{n}{p} ≤ 1,5$

$\Rightarrow p ≤ n ≤ 1,5p$

$\Rightarrow p ≤ 13 - 2p ≤ 1,5p$
$\Rightarrow 3,7 ≤ p ≤ 4,3$

Suy ra, với $p = 4$ thì thỏa mãn $\Rightarrow n = 13 - 2p = 5$

Vậy nguyên tử có 4 hạt proton, 4 hạt electron và 5 hạt notron

a)

$CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

$Cu(OH)_2 \xrightarrow{t^o} CuO + H_2O$

b)

Theo PTHH : 

$n_{CuO} = n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{300.40\%}{160} = 0,75(mol)$
$m_{CuO} = 0,75.80 = 60(gam)$

a) \(m_{CuSO_4}=300.40\%=120\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{120}{160}=0,75\left(mol\right)\)

PTHH:

`CuSO_4 + 2NaOH -> Cu(OH)_2 + Na_2SO_4`

$Cu(OH)_2 \xrightarrow{t^o} CuO +H_2O$

b) BTNT Cu: n_CuO = n_CuSO4 = 0,75 (mol)

=> m_CuO = 0,75.80 = 60 (g)

$n_{HCl} = 0,2.2 = 0,4(mol)$
$NaOH + HCl \to NaCl + H_2O$
Theo PTHH , $n_{HCl} = n_{NaOH} = n_{NaCl} = 0,4(mol)$

$\Rightarrow V_{dd\ NaOH} = \dfrac{0,4}{1,5} = \dfrac{4}{15}(lít)$

Sau phản ứng, $V_{dd} = \dfrac{4}{15} + 0,2 = \dfrac{7}{15}(lít)$
$\Rightarrow C_{M_{NaCl}} = \dfrac{0,4}{\dfrac{7}{15}} = 0,857M$

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