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\(n_{SO_2}=\dfrac{7.84}{22.4}=0.07\left(mol\right)\)
\(2SO_2+O_2\underrightarrow{^{t^0}}2SO_3\)
\(0.07.............0.07\)
\(m_{dd_{H_2SO_4}}=57.2\cdot1.5=85.8\left(g\right)\)
\(m_{H_2SO_4}=85.8\cdot60\%=51.48\left(g\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(0.07..................0.07\)
\(m_{dd}=0.07\cdot80+85.8=91.4\left(g\right)\)
\(\sum n_{H_2SO_4}=0.07\cdot98+51.48=58.34\left(g\right)\)
\(C\%_{H_2SO_4}=\dfrac{58.34}{91.4}\cdot100\%=63.8\%\)
\(n_{SO_3}=\dfrac{3}{80}=0,0375\left(mol\right)\\ pthh:SO_3+H_2O\rightarrow H_2SO_4\)
b) sản phẩm làm QT hóa đỏ vì sp là axit
\(pthh:SO_3+H_2O\rightarrow H_2SO_4\)
0,0375 0,0375
\(\Rightarrow m_{H_2SO_4}=0,0375.98=3,675\left(g\right)\)
\(C_M=\dfrac{0,0375}{0,25}=0,15M\)
a) \(n_{KMnO_4}=\dfrac{47,4}{158}=0,3\left(mol\right)\)
PTHH: 2KMnO4 --to--> K2MnO4+ MnO2 + O2
0,3------------------------------>0,15
=> VO2 = 0,15.22,4 = 3,36 (l)
b)
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
Xét tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,15}{5}\) => P hết, O2 dư
PTHH: 4P + 5O2 --to--> 2P2O5
0,1------------>0,05
P2O5 + 3H2O --> 2H3PO4
0,05-------------->0,1
=> \(C_{M\left(H_3PO_4\right)}=\dfrac{0,1}{0,08}=1,25M\)
a)2KMnO4→K2MnO4+MnO2+O2nKMnO4=47,4158=0,3molnO2=nKMnO42=0,15molVO2=0,15×22,4=3,36lb)O2+4Na→2Na2ONa2O+H2O→2NaOHnNa=9,223=0,4mol0,44<0,151⇒O2 dưnNa2O=nNa2=0,2molnNaOH=2nNa2O=0,4molCMNaOH=0,40,2=2Mc)nNaOH=0,1×2=0,2molVNaOH=nCM=0,20,5=0,4l=400mla)2KMnO4→K2MnO4+MnO2+O2nKMnO4=47,4158=0,3molnO2=nKMnO42=0,15molVO2=0,15×22,4=3,36lb)O2+4Na→2Na2ONa2O+H2O→2NaOHnNa=9,223=0,4mol0,44<0,151⇒O2 dưnNa2O=nNa2=0,2molnNaOH=2nNa2O=0,4molCMNaOH=0,40,2=2Mc)nNaOH=0,1×2=0,2molVNaOH=nCM=0,20,5=0,4l=400ml
Cách tiến hành : Cho thêm 300ml nước vào dung dịch A
Ta có: \(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)
a. PTHH: S + O2 ---to---> SO2
Theo PT: \(n_{SO_2}=n_S=0,2\left(mol\right)\)
=> \(m_{SO_2}=0,2.64=12,8\left(g\right)\)
b. Theo PT: \(n_{O_2}=n_S=0,2\left(mol\right)\)
=> \(m_{O_2}=0,2.32=6,4\left(g\right)\)
a)S+O2-------->SO2
b)n S=6,4/32=0,2(mol)
Theo pthh
n SO2 =n S=0,2(mol)
V SO2=0,2.22,4=4,48(mol)
a, PTHH: S + O2 -> (t°) SO2
b, nS = 6,4/32 = 0,2 (mol)
nO2 = 6,72/22,4 = 0,3 (mol)
LTL: 0,2 < 0,3 => O2 dư
nO2 (pư) = nSO2 = nS = 0,2 (mol)
mO2 (dư) = (0,3 - 0,2) . 32 = 3,2 (g)
c, mSO2 = 64 . 0,2 = 12,8 (g)
a, \(S+O_2\underrightarrow{t^o}SO_2\)
\(nS=\dfrac{6,4}{32}=0,2\left(mol\right)\)
\(nO_2=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => oxi dư
\(nO_{2\left(dư\right)}=0,1\left(mol\right)\)
\(mO_{2\left(dư\right)}=0,1.32=3,2\left(g\right)\)
\(nSO_2=nS=0,2\left(mol\right)\)
\(mSO_2=0,2.64=12,8\left(g\right)\)
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)
c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4\left(LT\right)}=2n_{O_2}=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow m_{KMnO_4\left(LT\right)}=\dfrac{2}{15}.158=\dfrac{316}{15}\left(g\right)\)
Mà: H% = 85%
\(\Rightarrow m_{KMnO_4\left(TT\right)}=\dfrac{\dfrac{316}{15}}{85\%}\approx24,78\left(g\right)\)
nH2SO4 = 0,2 mol
SO3 + H2O \(\rightarrow\) H2SO4
\(\Rightarrow\) nSO3 = 0,2 mol
2SO2 + O2 \(\xrightarrow[V_2O_5]{t^o}\) 2SO3
\(\Rightarrow\) nSO2 = 0,2 mol
S + O2 \(\underrightarrow{t^o}\) SO2
\(\Rightarrow\) nS = 0,2 mol
\(\Rightarrow\) mS thực = 0,2.\(\dfrac{100}{80}\).32 = 8 (g)
nH2SO4 = 19.6/98=0.2mol
SO3 + H2O -> H2SO4
(mol) 0.2 0.2
2SO2 + O2 -> 2SO3
(mol) 0.2 0.2
S + O2 -> SO2
(mol) 0.2 0.2
mSO2 = 0.2*64=12.8g
Khối luowjng SO2 thực: 12.8*100/80=16g