nS=0.5(mol)

S+O2->SO2

Theo pthh nS=nO2=nSO2->nS=nO2=nSO2=0.5(mol)

->V1=V2=11.2(l)

PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)

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

Ta có: \(n_{KClO_3}=\dfrac{18,375}{122,5}=0,15\left(mol\right)\) \(\Rightarrow n_{O_2\left(lý.thuyết\right)}=0,225\left(mol\right)\)

\(\Rightarrow n_{O_2\left(thực\right)}=0,225\cdot85\%=0,19125\left(mol\right)=n_S=n_{SO_2}\)

\(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,19125\cdot22,4=4,284\left(l\right)=V_{SO_2}\\m_S=0,19125\cdot32=6,12\left(g\right)\\\end{matrix}\right.\)

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

Theo PTHH: \(V_{O_2}=22,4\cdot20\%=4,48\left(l\right)=V_{SO_2}\) 

\(\Rightarrow n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_S\) \(\Rightarrow m_S=0,2\cdot32=6,4\left(g\right)\)

nS=0,2mol

nO2=0,35mol

PTHH: S+O2=>SO2

             0,2:0,35

=> nO2 dư theo nS

p/ư:       0,2->0,2->0,2

=> V =0,2.22,4=4,48l

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

\(nO_2=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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

1      1       1 (mo)

0,2  0,2   0,2

LTL : 0,2 / 1  < 0,3 / 1

=> S đủ , O2 dư

= > mSO2 = 0,2 . 64 = 12, 8 (g)

=> Chọn C

Đáp án là D. 16,8

2Cu+O2-to>2CuO

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

4P+5O2-to>2P2O5

n Cu=\(\dfrac{6,4}{64}\)=0,1 mol

=>VO2=0,05.22,4=1,12l

=>m CuO=0,1.80=8g

b) 

thiếu đề

a. \(n_{Cu}=\dfrac{6.4}{64}=0,1\left(mol\right)\)

PTHH : 2Cu + O₂ -> 2CuO

            0,1       0,05     0,1

\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)

\(m_{CuO}=0,1.81=8,1\left(g\right)\)

b. Thiếu số mol P 

\(n_{SO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

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

\(n_S=0.1\left(mol\right)\)

\(m_S=0.1\cdot32=3.2\left(g\right)\)

=> A 

PTHH : S + O2 -> SO2

nSO2 = V/22,4= 0,1 mol

Theo PTHH : nS = nSO2 = 0,1 mol

=> mS = n.M = 3,2 g

Theo gt ta có: $n_{O_2}=0,15625(mol);n_{SO_2}=0,1(mol)$

$S+O_2\rightarrow SO_2$

Ta có: $n_{S}=0,1(mol)\Rightarrow m_{S}=3,2(g)$