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`a)PTHH:`
`Zn + 2HCl -> ZnCl_2 + H_2`
`0,02` `0,02` `0,02` `(mol)`
`n_[Zn]=[1,3]/65=0,02(mol)`
`b)V_[H_2]=0,02.22,4=0,448(l)`
`c)C%_[ZnCl_2]=[0,02.136]/[1,3+50-0,02.2].100~~5,31%`
\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,02 0,04 0,02 0,02 ( mol )
\(V_{H_2}=0,02.22,4=0,448\left(l\right)\)
\(C\%_{ZnCl_2}=\dfrac{0,02.136}{1,3+50-0,02.2}.100=5,3\%\)
a. Ta có: mNaOH=\(\frac{200.20}{100}=40\left(g\right)\)
pt : NaOH + HCl --------> NaCl + H2O
theo pt: 40g 36,5g 58,5g 18g
theo đề: 40g 36,5g 58,5g
=>\(C_{\%}=\frac{58,5}{200+100}.100\%=19,5\%\)
b.\(C_{\%}=\frac{36,5}{100}.100\%=36,5\%\)
a)Khối lượng của dd NaOH:
\(m_{NaOH}=\dfrac{m_{dd}.C\%}{100\%}=\dfrac{400.30\%}{100\%}=120\left(g\right)\)
Số mol của 400 g dd NaOH:
\(n_{NaOH}=\dfrac{m}{M}=\dfrac{120}{40}=3\left(mol\right)\)
PTHH:
\(NaOH+HCl\rightarrow NaCl+H_2O\)
1 : 1 : 1 : 1
3 -> 3 : 3 : 3 (mol)
Khối lượng của 3 mol NaCl:
\(m_{NaCl}=n.M=3.36,5=109,5\left(g\right)\)
Khối lượng của dd NaCl sau P.Ư:
\(m_{ddNaCl}=m_{ct}+m_{dm}=100+400=500\left(g\right)\)
Nồng độ % của dd NaCl:
\(C\%_{NaCl}=\dfrac{m_{ct}}{m_{dm}}.100\%=\dfrac{3.58,5}{500}.100\%=35,1\%\)
b, Nồng độ phần trăm của Axit:
\(C\%_{HCl}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{3.36,5}{500}.100\%=21,9\%\)
\(n_{FeO}=\dfrac{10.8}{72}=0.15\left(mol\right)\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
\(0.15.......0.3.............0.15\)
\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)
\(C\%HCl=\dfrac{10.95}{100}\cdot100\%=10.95\%\)
\(m_{dd}=10.8+100=110.8\left(g\right)\)
\(m_{FeCl_2}=0.15\cdot127=19.05\left(g\right)\)
\(C\%FeCl_2=\dfrac{19.05}{110.8}\cdot100\%=17.19\%\)
nFeO=10.872=0.15(mol)nFeO=10.872=0.15(mol)
FeO+2HCl→FeCl2+H2OFeO+2HCl→FeCl2+H2O
0.15.......0.3.............0.150.15.......0.3.............0.15
mHCl=0.3⋅36.5=10.95(g)mHCl=0.3⋅36.5=10.95(g)
C%HCl=10.95100⋅100%=10.95%C%HCl=10.95100⋅100%=10.95%
mdd=10.8+100=110.8(g)mdd=10.8+100=110.8(g)
mFeCl2=0.15⋅127=19.05(g)mFeCl2=0.15⋅127=19.05(g)
C%FeCl2=19.05110.8⋅100%=17.19%C%FeCl2=19.05110.8⋅100%=17.19%
\(n_{NaOH}=\dfrac{20\%.200}{40}=1\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaCl}=n_{HCl}=n_{NaOH}=1\left(mol\right)\\ a,m_{ddNaCl}=200+100=300\left(g\right)\\ C\%_{ddNaCl}=\dfrac{58,5.1}{300}.100=19,5\%\\ b,C\%_{ddHCl}=\dfrac{36,5.1}{100}.100=36,5\%\)
`a)PTHH:`
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`0,2` `0,6` `0,3` `(mol)`
`n_[Al]=[5,4]/27=0,2(mol)`
`b)V_[H_2]=0,3.22,4=6,72(l)`
`c)m_[dd HCl]=[0,6.36,5]/10 . 100 =219(g)`
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
_____0,2____0,4_____0,2___0,2 (mol)
b, Ta có: m dd sau pư = mZn + m dd HCl - mH2 = 13 + 100 - 0,2.2 = 112,6 (g)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,2.136}{112,6}.100\%\approx24,16\%\)
c, Ta có: mHCl = 0,4.36,5 = 14,6 (g)
\(\Rightarrow C\%_{HCl}=\dfrac{14,6}{100}.100\%=14,6\%\)
Bạn tham khảo nhé!
a, PTHH: Zn + 2HCl ➝ ZnCl2 + H2
(mol) 1 2 1 1
(mol) 0.2
b, nZn=13 :65 =0.2 (mol)
Theo PTHH: nZnCl2=(0.2x1):1=0.2(mol)
→mZnCl2=0.2x(65+2x35.5)=27.2(g)
⇒C%ZnCl2=27.2:100x100=27.2(%)
c,Theo PTHH: nHCl =(0.2 x 2) :1=0.4(mol)
➝mHCl=0.4x(1+35.5)=14.6(g)
⇒C%HCl=14.6:100x100%=14.6(%)