`#Ly`

$a)$

$n_{O_2}$ $=$ $\dfrac{V}{22,4}$ $=$ $\dfrac{13,44}{22,4}$ $=$ $0,6$ $(mol)$

Giả sử $x,y$ là số $mol$ $S,P$ phản ứng $(x,y>0)$

PTHH:

$(1):$ $S$ $+$ $O_2$ $\xrightarrow[]{t^o}$ $SO_2$

$(2):$ $4P$ $+$ $5O_2$ $\xrightarrow[]{t^o}$ $2P_2O_5$

Theo PT $(1)$, $n_{O_2}$ $=$ $n_{S}$ $=$ $x$ $(mol)$

Theo PT $(2)$, $n_{O_2}$ $=$ $\dfrac{5}{4}n_{P}$ $=$ $1,25y$ $(mol)$

$⇒$ $\sum\limits_{}^{}$$n_{O_2}$ $=$ $x+1,25y$ $(mol)$

$m_{S,P}$ $=$ $32x+31y$ $(g)$

Ta có hệ PT sau:

$x+1,25y=0,6$

$32x+31y=15,6$

$⇒$ $\begin{cases} x=0,1\\y=0,4 \end{cases}$

$m_{S}$ $=$ $n.M=0,1.32=3,2$ $(g)$

`%m_(S` $=$ $\dfrac{m_{S}}{m_{hh}}$`.100%` $=$ $\dfrac{3,2}{15,6}$`.100%` $≈$ `20,5%`

`%m_(P` $=$ $100-20,5=79,5$`%`

$b)$

$n_{Cu}$ $=$ $\dfrac{m}{M}$ $=$ $\dfrac{25,6}{64}$ $=$ $0,4$ $(mol)$

PTHH: $2Cu$ $+$ $O_2$ $\xrightarrow[]{t^o}$ $2CuO$

Ta xét tỉ lệ: $\dfrac{0,4}{2}$ $<$ $\dfrac{0,6}{1}$

$⇒$ $Cu$ pư hết, $O_2$ dư nên tính theo $n_{Cu}$

Theo PT trên, $n_{CuO}$ $=$ $n_{Cu}$ $=$ $0,4$ $(mol)$

$m_{CuO}$ $=$ $n.M=0,4.80=32$ $(g)$

Theo PT, $n_{O_2(pư)}$ $=$ $\dfrac{1}{2}n_{Cu}$ $=$ $0,2$ $(mol)$

$n_{O_2(dư)}$ $=$ $0,6-0,2=0,4$ $(mol)$

$m_{O_2(dư)}$ $=$ $n.M=0,4.32=12,8$ $(g)$

\(\begin{array}{l}
a)\\
Dat\,n_S=x(mol);n_P=y(mol)\\
\to 32x+31y=15,6(1)\\
n_{O_2}=\frac{13,44}{22,4}=0,6(mol)\\
S+O_2\xrightarrow{t^o}SO_2\\
4P+5O_2\xrightarrow{t^o}2P_2O_5\\
Theo\,PT:\,x+\frac{5}{4}y=n_{O_2}=0,6(2)\\
Tu\,(1)(2)\to\begin{cases} x=0,1\\ y=0,4 \end{cases}\\
\to \%m_S=\frac{0,1.32}{15,6}.100\%≈20,5\%\\
\to \%m_P=100-20,5=79,5\%\\
b)\\
n_{Cu}=\frac{25,6}{64}=0,4(mol)\\
2Cu+O_2\xrightarrow{t^o}2CuO\\
Vi\,\frac{n_{Cu}}{2}=0,2<0,6=n_{O_2}\to O_2\,du\\
Theo\,PT:\,n_{CuO}=n_{Cu}=0,4(mol)\\
\to m_{CuO}=0,4.80=32(g)
\end{array}\)