\(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

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

0,2 0,5 ---> 0,2

Lập tỉ số: \(n_S:n_{O_2}=0,2< 0,5\)

S hết, O_2dư

_{\(V_{O_2\left(dư\right)}=\left(0,5-0,2\right).22,4=6,72\left(l\right)\)}

_{\(V_{SO_2}=0,2.22,4=4,48\left(l\right)\)}

PTHH: S + O₂ \(\underrightarrow{t^o}\) SO₂\(\uparrow\)

Theo pt: 1 ... 1 ....... 1 .. (mol)

Theo đề: 0,2 0,2 ... 0,2 .. (mol)

\(n_S=\dfrac{m}{M}=\dfrac{6,4}{32}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{V_{đktc}}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

So sánh \(\dfrac{n_{S_{đề}}}{n_{S_{pt}}}< \dfrac{n_{O_{2_{đề}}}}{n_{O_{2_{pt}}}}\left(0,2< 0,5\right)\)

=> O₂ dư, tính theo ⁿS

=> Các khí sinh ra sau phản ứng gồm SO₂ và O₂ dư.

\(V_{SO_{2_{đktc}}}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(V_{O_{2\left(dư\right)_{đktc}}}=n.22,4=\left(0,5-0,2\right).22,4=6,72\left(l\right)\)

\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)

\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(0.05.0.05...0.05\)

\(\Rightarrow Sdư\)

\(V_{SO_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(b.\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(0.1..0.1\)

\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)

a, PT: \(S+O_2\underrightarrow{t^o}SO_2\)

Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được S dư.

Theo PT: \(n_{SO_2}=n_{O_2}=0,05\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

b, Theo PT: \(n_{O_2}=n_S=0,1\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

\(\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)

Câu 11: 

\(n_{CaO}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

PTHH:           \(CaO+H_2SO_4\rightarrow CaSO_4+H_2O\)

Ban đầu:         0,2       0,4             0,2

Sau pư:            0         0,2            0,2

`=>`\(\left\{{}\begin{matrix}m_{H_2SO_4}=0,2.98=19,6\left(g\right)\\m_{CaSO_4}=0,2.136=27,2\left(g\right)\end{matrix}\right.\)

Câu 12:

\(n_S=\dfrac{6,4}{32}=0,2\left(mol\right);n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:           \(S+O_2\xrightarrow[]{t^o}SO_2\)

Ban đầu:       0,2   0,5

Sau pư:         0     0,3       0,2

`=>`\(\left\{{}\begin{matrix}V_{O_2}=0,3.22,4=6,72\left(l\right)\\V_{SO_2}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

Câu 13:

\(n_C=\dfrac{4,8}{12}=0,4\left(mol\right);n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH:           \(C+O_2\xrightarrow[]{t^o}CO_2\)

Ban đầu:      0,4   0,3

Sau pư:        0,1   0        0,3

`=>`\(\left\{{}\begin{matrix}m_{C\left(d\text{ư}\right)}=0,1.12=1,2\left(g\right)\\V_{CO_2}=0,3.22,4=6,72\left(l\right)\end{matrix}\right.\)

Câu 14:

\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right);n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PTHH:         \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)

Ban đầu:       0,1           0,1

Sau pư:         0              0               0,1           0,2

`=>`\(\left\{{}\begin{matrix}m_{BaSO_4}=0,1.233=23,3\left(g\right)\\m_{HCl}=0,2.36,5=7,3\left(g\right)\end{matrix}\right.\)

Câu 15: 

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH:          \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

Ban đầu:        0,25     0,5

Sau pư:           0          0            0,25

`=>`\(m_{CuCl_2}=0,25.135=33,75\left(g\right)\)

ssc là ?

`S + O_2` $\xrightarrow{t^o}$ `SO_2`

        `0,5`        `0,5`                   `(mol)`

`n_[SO_2]=[11,2]/[22,4]=0,5(mol)`

`=>m_[O_2]=0,5.32=16(g)`