26 chia 2 bằng 13 mằ :v

Có: \(\left\{{}\begin{matrix}2p+n=40\\2p-n=12\end{matrix}\right.\left(do:e=p\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}p=e=13\\n=14\end{matrix}\right.\)

`1)`

Ta có: \(\left\{{}\begin{matrix}p+e+n=49\\n=53,125\%.\left(p+e\right)\end{matrix}\right.\)

Do `p = n`

\(\Rightarrow\left\{{}\begin{matrix}2p+n=49\\n-53,125\%.2p=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}p=e=16\\n=17\end{matrix}\right.\)

`2)` Lưu huỳnh - `S`

Giả sử có 100g dd H₂SO₄

\(\Rightarrow m_{H_2SO_4}=\dfrac{9,8}{100}.100=9,8\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PTHH:

               \(M_2\left(CO_3\right)_n+nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nCO_2\uparrow+nH_2O\)

               \(\dfrac{0,1}{n}\)<----------0,1------------->\(\dfrac{0,1}{n}\)----------->0,1

\(\Rightarrow\left\{{}\begin{matrix}m_{M_2\left(CO_3\right)_n}=\dfrac{0,1}{n}.\left(2M_M+60n\right)\left(g\right)\\m_{M_2\left(SO_4\right)_n}=\dfrac{0,1}{n}.\left(2M_M+96n\right)\left(g\right)\\m_{CO_2}=0,1.44=4,4\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{\dfrac{0,1}{n}.\left(2M_M+96n\right)}{\dfrac{0,1}{n}.\left(2M_M+60n\right)+100-4,4}.100\%=14,45\%\)

\(\Rightarrow M_M=30n\left(g\text{/}mol\right)\)

Xét các giá trị 1, 2, 3 thì không có giá trị nào tm nên đề sai

uk

$n_{O_2(b đ)}  = \dfrac{33,6}{22,4} = 1,5 (mol)$

PTHH:

$2Zn + O_2 \xrightarrow{t^o} 2ZnO$

0,5---->0,25--->0,5

$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

0,2---->0,15--->0,1

$\rightarrow n_{O_2(cần)} = 0,25 + 0,15 = 0,4 (mol)$

Xét `0,4 < 1,5 => O_2` dư

$m_{c rắn} = 0,5.81 + 0,1.102 = 50,7 (g)$

`a)`

$n_{Al} = \dfrac{10,8}{27} = 0,4 (mol)$

PTHH: $4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

Theo PT: $n_{Al_2O_3} = \dfrac{1}{2} n_{Al} = 0,2 (mol)$

$\rightarrow m_{Al_2O_3(tt)} = 0,2.75\%.10 = 15,3 (g)$

$b)$

$m_{Al(dư)} = 0,4 - 0,4.75\% = 0,1 (mol)$

$\rightarrow m = 0,1.27 + 15,3 = 18 (g)$

`a)`

\(n_{FeS}=\dfrac{3,52}{88}=0,04\left(mol\right)\)

Trong 22,4 lít không khí có: \(\left\{{}\begin{matrix}V_{O_2}=\dfrac{20}{100}.22,4=4,48\left(l\right)\\V_{N_2}=22,4-4,48=17,92\left(l\right)\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{N_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\end{matrix}\right.\)

PTHH: 

                      \(4FeS+7O_2\xrightarrow[]{t^o}2Fe_2O_3+4SO_2\uparrow\)

ban đầu         0,04      0,2

sau pư            0         0,13       0,02           0,04

`=> m_A = m_{Fe_2O_3} = 0,02.160 = 3,2 (g)`

`b)`

Trong khí B có \(\left\{{}\begin{matrix}SO_2:0,04mol\\O_2:0,13mol\\N_2:0,8\left(mol\right)\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}n_B=0,04+0,13+0,8=0,97\left(mol\right)\\m_B=0,04.64+0,13.32+0,8.28=29,12\left(g\right)\end{matrix}\right.\)

$\Rightarrow M_B = \dfrac{29,12}{0,97} = 30,02 (g/mol)$

$\Rightarrow d_{B/H_2} = \dfrac{30,02}{2} = 15,01$