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\(n_{Na_2CO_3}=n_{Na_2CO_3\cdot10H_2O}=\dfrac{57.2}{106+18\cdot10}=0.2\left(mol\right)\)
\(C_{M_{Na_2CO_3}}=\dfrac{0.2}{0.4}=0.5\left(M\right)\)
\(m_{Na_2CO_3}=0.2\cdot106=21.2\left(g\right)\)
\(m_{dd}=400\cdot1.05=420\left(g\right)\)
\(C\%_{Na_2CO_3}=\dfrac{21.2}{420}\cdot100\%=5.04\%\)
a) \(m_{HCl}=200\cdot7,3\%=14,6\left(g\right)\)
b) \(n_{NaOH}=0,5\cdot1=0,5\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,5\cdot40=20\left(g\right)\)
c) \(n_{CuSO_4}=0,2\cdot1,5=0,3\left(mol\right)\) \(\Rightarrow m_{CuSO_4}=0,3\cdot160=48\left(g\right)\)
d) Bạn xem lại đề !
a) mHCl=200⋅7,3%=14,6(g)mHCl=200⋅7,3%=14,6(g)
b) nNaOH=0,5⋅1=0,5(mol)nNaOH=0,5⋅1=0,5(mol) ⇒mNaOH=0,5⋅40=20(g)⇒mNaOH=0,5⋅40=20(g)
c) nCuSO4=0,2⋅1,5=0,3(mol)nCuSO4=0,2⋅1,5=0,3(mol) ⇒mCuSO4=0,3⋅160=48(g)⇒mCuSO4=0,3⋅160=48(g)
d) Bạn xem lại đề !
Zn+H2SO4->ZnSO4+H2
0,11-------------------------0,11
2KMnO4-tO>K2MnO4+MnO2+O2
0,06-------------------------------------0,03
2H2+O2-to>2H2O
0,06---0,03-0,06
n Zn=0,11 mol
n KMnO4=0,06 mol
=>H2 du2
=>m H2O=0,06.18=1,08g
\(n_{Zn}=\dfrac{7,15}{65}=0,11\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,11 0,11
\(n_{KMnO_4}=\dfrac{9,48}{158}=0,06\left(mol\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,06 0,03
\(pthh:2H_2+O_2\underrightarrow{t^o}2H_2O\\
LTL:\dfrac{0,11}{2}>\dfrac{0,03}{1}=>H_2d\text{ư}\)
theo pthh : nH2O =2 nO2= 0,12 (mol)
=> mH2O = 0,12 . 18 = 2,16(g)
1
\(a)m_{H_2O}=250-5=245g\\b )C_{\%NaCl}=\dfrac{5}{250}\cdot100=2\%\)
\(2\\ m_{ddCuSO_4}=\dfrac{15.100}{5}=300g\\ m_{H_2O}=300-15=285g\)
Câu 1:
a, Ta có: m dd = m chất tan + mH2O ⇒ mH2O = 250 - 5 = 245 (g)
b, \(C\%_{NaCl}=\dfrac{5}{250}.100\%=2\%\)
Câu 2:
Ta có: \(C\%_{CuSO_4}=\dfrac{15}{m_{ddCuSO_4}}.100\%=5\%\)
\(\Rightarrow m_{ddCuSO_4}=300\left(g\right)\)
⇒ mH2O = 300 - 15 = 285 (g)
\(n_{Al}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PTHH: Al2O3 + 6HNO3 ---> 2Al(NO3)3 + 3H2O
0,1 0,6 0,2 0,3
\(\rightarrow m_{HNO_3}=0,6.63=37,8\left(g\right)\\ m_{ddHNO_3}=\dfrac{37,8}{15\%}=252\left(g\right)\\ m_{dd\left(sau.pư\right)}=252+10,2=262,2\left(g\right)\\ m_{Al\left(NO_3\right)_3}=0,2.213=42,6\left(g\right)\\ C\%_{Al\left(NO_3\right)_3}=\dfrac{42,6}{262,2}=16,25\%\)
\(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)
\(Al_2O_3+6HNO_3\rightarrow2Al\left(NO_3\right)_3+3H_2O\)
\(0.1...........0.6..................................0.3\)
\(m_{dd_{HNO_3}}=\dfrac{0.6\cdot63\cdot100}{15}=252\left(g\right)\)
Số phân tử nước :
\(0.3\cdot6\cdot10^{23}=1.8\cdot10^{23}\left(pt\right)\)
