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a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Có lẽ phần này đề hỏi khối lượng sắt chứ bạn nhỉ?
\(n_{ZnCl_2}=0,4.2=0,8\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{FeCl_2}=0,8\left(mol\right)\Rightarrow m_{Fe}=0,8.56=44,8\left(g\right)\)
c, \(n_{H_2}=n_{FeCl_2}=0,8\left(mol\right)\Rightarrow V_{H_2}=0,8.22,4=17,92\left(l\right)\)
d, \(n_{HCl}=2n_{FeCl_2}=1,6\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{1,6}{0,4}=4\left(M\right)\)
a) \(PT:CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
\(HCl+NaOH\rightarrow NaOH+H_2O\)
b) \(m_{HCl}=\frac{200.10,95\%}{100\%}=21,9\left(g\right)\)
\(n_{HCl}=\frac{21,9}{36,5}=0,6\left(mol\right)\)
c) \(n_{NaOH}=2.0,05=0,1\left(mol\right)\Rightarrow n_{HCl\left(pưNaOH\right)}=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl\left(pưCaCO_3\right)}=0,6-0,1=0,5\left(mol\right)\)
d) \(n_{CaCO_3}=\frac{1}{2}n_{HCl\left(pưCaCO_3\right)}=0,5.\frac{1}{2}=0,25\left(mol\right)\)
\(m_{CaCO_3}=0,25.100=25\left(g\right)\)
e) \(n_{CO_2}=n_{CaCO_3}=0,25\left(mol\right)\)
\(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)
f) \(n_{CaCl_2}=n_{CaCO_3}=0,25\left(mol\right)\)
\(m_{ddA}=25+200-0,25.44=214\left(g\right)\)
\(C\%_{ddCaCl_2}=\frac{0,25.111}{214}.100\%=12,97\%\)
\(C\%_{ddHCldư}=\frac{0,1.36,5}{214}.100\%=1,71\%\)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,1-->0,2------>0,1-->0,1
=> VH2 = 0,1.22,4 = 2,24 (l)
mZnCl2 = 0,1.136 = 13,6 (g)
b) \(C\%_{dd.HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)
c) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,15}{1}\) => H2 hết, O2 dư
PTHH: 2H2 + O2 --to--> 2H2O
0,1--------------->0,1
=> mH2O = 0,1.18 = 1,8 (g)
a)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4----->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) mHCl = 0,4.36,5 = 14,6 (g)
=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)
c)
mdd sau pư = 13 + 200 - 0,2.2 = 212,6 (g)
mZnCl2 = 0,2.136 = 27,2 (g)
=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)
Câu 1:
\(\text{a) }pthh:CaCO3+2HCl\rightarrow CaCl_2+CO_2+H_2O\left(1\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\left(2\right)\)
b) \(n_{NaOH}=C_M\cdot V=0,05\cdot2=0,1\left(mol\right)\)
Theo \(pthh\left(2\right):n_{HCl\left(2\right)}=n_{NaOH}=0,1\left(mol\right)\)
\(m_{HCl}=\dfrac{m_{d^2HCl}\cdot C\%}{100}=\dfrac{200\cdot10,95}{100}=21,9\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{m}{M}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\ \Rightarrow n_{HCl\left(1\right)}=0,6-0,1=0,5\left(mol\right)\)
Theo \(pthh\left(1\right):n_{CaCO_3}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=n\cdot M=0,25\cdot100=25\left(g\right)\)
c) Theo \(pthh\left(1\right):n_{CO_2}=\dfrac{1}{2}n_{HCl\left(1\right)}=\dfrac{1}{2}\cdot0,5=0,25\left(mol\right)\)
\(\Rightarrow V_{CO_2}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\)
d) \(m_{CO_2}=n\cdot M=0,25\cdot44=11\left(g\right)\)
\(m_{HCl\left(dư\right)}=n\cdot M=0,1\cdot36,5=3,65\left(g\right)\)
\(m_{d^2A}=\left(m_{CaCO_3}+m_{d^2HCl}\right)-m_{CO_2}\\ =\left(25+200\right)-11=214\left(g\right)\)
\(\Rightarrow C\%\left(HCl_{dư}\right)=\dfrac{3,65\cdot100}{214}=1,71\%\)
a) \(m_{NaOH}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{200\cdot15}{100}=30\left(g\right)\)
\(m_{d^2A\text{ sau khi pha thêm }100\left(g\right)nước}=100+200=300\left(g\right)\)
\(\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{30\cdot100}{300}=10\%\)
b) \(m_{NaOH\text{ sau khi cho thêm }5\left(g\right)NaOH}=30+5=35\left(g\right)\)
\(m_{d^2B}=5+200=205\left(g\right)\)
\(\Rightarrow C\%\left(NaOH\right)=\dfrac{m_{NaOH}\cdot100}{m_{d^2}}=\dfrac{35\cdot100}{205}=17,07\%\)
c) \(C\%\left(NaOH\right)=\dfrac{30\cdot100}{150}=20\%\)
a, PTPƯ: SO3 + H2O ---> H2SO4
nSO3=\(\dfrac{2,24}{22,4}=0,1mol\)
1 mol SO3 ---> 0,1 mol H2SO4
nên 0,1 mol SO3 ---> 0,1 mol H2SO4
CM H2SO4=\(\dfrac{0,1}{0,5}\)=0,2 M
b, PTPƯ: Zn + H2SO4 ---> ZnSO4 + H2
1 mol H2SO4 ---> 1 mol Zn
nên 0,1 mol H2SO4 ---> 0,1 mol Zn
mZn=0,1.65=6,5 g
a) \(m_{HCl}=\dfrac{200.10,95}{100}=21,9\left(g\right)\)
=> \(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
b) \(n_{CaCO_3}=\dfrac{a}{100}=0,01a\left(g\right)\)
\(n_{NaOH}=0,05.2=0,1\left(mol\right)\)
PTHH: CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
______0,01a---->0,02a---->0,01a->0,01a___________(mol)
NaOH + HCl --> NaCl + H2O
_0,1----->0,1___________________________________(mol)
=> 0,02a = 0,6 - 0,1
=> a = 25 (g)
c) \(V_{CO_2}=0,01.25.22,4=5,6\left(l\right)\)
d) \(\left\{{}\begin{matrix}C\%\left(CaCl_2\right)=\dfrac{0,25.111}{25+200-0,25.44}.100\%=12,97\%\\C\%\left(HCl_{dư}\right)=\dfrac{0,1.36,5}{25+200-0,25.44}.100\%=1,705\%\end{matrix}\right.\)