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\(n_{SO_2}=\dfrac{V_{SO_2\left(ĐKTC\right)}}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(S+O_2\underrightarrow{t^o}SO_2\)
...........1.........1........1......
...........0,3......0,3......0,3.....
a. \(m_S=n_S\cdot M_S=0,3\cdot32=9,6\left(g\right)\)
b. \(V_{O_2\left(ĐKTC\right)}=n_{O_2}\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)
\(V_{kk\left(ĐKTC\right)}=V_{O_2\left(ĐKTC\right)}\cdot5=6,72\cdot5=33,6\left(l\right)\)
a)
\(m_{MgCl_2}=\dfrac{50.4}{100}=2\left(g\right)\Rightarrow m_{H_2O}=50-2=48\left(g\right)\)
b)
\(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)
PTHH: S + O2 --to--> SO2
0,2->0,2
=> VO2 = 0,2.22,4 = 4,48 (l)
=> Vkk = 4,48 : 20% = 22,4 (l)
a.\(m_{MgCl_2}=\dfrac{50.4}{100}=2g\)
\(m_{H_2O}=50-2=48g\)
b.\(n_S=\dfrac{6,4}{32}=0,2mol\)
\(S+O_2\rightarrow\left(t^o\right)SO_2\)
0,2 0,2 ( mol )
\(V_{kk}=V_{O_2}.5=\left(0,2.22,4\right).5=22,4l\)
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) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,3------------------>0,15----->0,45
=> \(m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)
b)
PTHH: 2H2 + O2 --to-->2H2O
0,45->0,225
=> \(V_{O_2}=0,225.22,4=5,04\left(l\right)\)
=> Vkk = 5,04 : 20% = 25,2 (l)
\(n_{hhkhí}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
Gọi \(n_{SO_2}=a\left(mol\right)\left(0< a< 0,75\right)\)
\(\rightarrow n_{O_2\left(dư\right)}=0,75-b\left(mol\right)\)
Ta có: \(\dfrac{64a+32\left(0,75-a\right)}{0,75}=\dfrac{33,6}{1}=33,6\left(\dfrac{g}{mol}\right)\)
\(\rightarrow a=0,0375\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%V_{SO_2}=\dfrac{0,0375}{0,75}=5\%\\\%V_{O_2\left(dư\right)}=100\%-5\%=95\%\end{matrix}\right.\)
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
`=>`\(m_{CuCl_2}=0,25.135=33,75\left(g\right)\)
\(n_S=\dfrac{6.4}{32}=0.2\left(mol\right)\)
\(S+O_2\underrightarrow{^{^{t^o}}}SO_2\)
\(0.2....0.2.....0.2\)
\(m_{SO_2}=0.2\cdot64=12.8\left(g\right)\)
\(V_{kk}=5V_{O_2}=5\cdot0.2\cdot22.4=22.4\left(l\right)\)
So mol cua luu huynh
nS = \(\dfrac{m_S}{M_S}=\dfrac{6,4}{32}=0,2\) (mol)
Pt : S + O2 \(\rightarrow\) SO2\(|\)
1 1 1
0,2 0,2 0,2
a) So mol cua luu huynh dioxit
nSO2 = \(\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
Khoi luong cua luu huynh dioxit
mSO2 = nSO2 . MSO2
= 0,2 . 64
= 12,8(g)
b) So mol cua khi oxi
nO2 = \(\dfrac{0,2.1}{1}=0,2\) (mol)
The tich cua khi oxi o dktc
VO2 = nO2 .22,4
= 0,2 .22,4
= 4,48(l)
The tich cua khong khi
VO2 = \(\dfrac{1}{5}\) Vkk \(\Rightarrow\) Vkk = 5 . VO2
= 5 . 4,48
= 22,4 (l)
Chuc ban hoc tot
\(n_{Al}=\dfrac{18,9}{27}=0,7\left(mol\right)\\ pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
0,7 0,525 0,35
\(V_{O_2}=\left(0,525.24\right)=12,6l\)
\(2Al+3S\underrightarrow{t^o}Al_2S_3\)
0,7 0,35
\(m_{Al_2S_3}=0,35.150=52,5g\)