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a) Đề bài sai phải không? (CO3 ----> CO2 chứ)
CO2 + H2O -----> H2CO3
CM= 0,36 (M)
b) nCuO=\(\frac{10}{80}\) = 0,125 (mol)
nH2CO3= \(\frac{2}{11}\) (mol)
CuO + H2CO3 ------> CuCO3 + H2O
ban đầu 0,125 \(\frac{2}{11}\) }
pư 0,125 ---> 0,125 ---> 0,125 } (mol)
sau pư 0 \(\frac{5}{88}\) 0,125 }
CM(H2CO3)=\(\frac{\frac{5}{88}}{0,5}\)=0,11 (M)
CM(CuCO3)=\(\frac{0,125}{0,5}\)=0,25 (M)
a) PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)
b) Ta có: \(n_{KCl}=0,15\cdot0,5=0,075\left(mol\right)=n_{KOH}\) \(\Rightarrow m_{KOH}=0,075\cdot56=4,2\left(g\right)\)
c) PTHH: \(KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\)
Theo PTHH: \(n_{KCl}=0,075\left(mol\right)=n_{AgNO_3\left(p.ứ\right)}=n_{KNO_3}=n_{AgCl}\)
\(\Rightarrow n_{AgNO_3\left(dư\right)}=0,075\cdot120\%-0,075=0,015\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{AgCl}=0,075\cdot143,5=10,7625\left(g\right)\\C_{M_{KNO_3}}=\dfrac{0,075}{0,5+2}=0,03\left(M\right)\\C_{M_{AgNO_3\left(dư\right)}}=\dfrac{0,015}{2,5}=0,006\left(M\right)\end{matrix}\right.\)
d) Coi như khi cô cạn không bị hao hụt muối
Ta có: \(m_{muối.khan}=m_{KNO_3}+m_{AgNO_3\left(dư\right)}=0,075\cdot101+0,015\cdot170=10,125\left(g\right)\)
\(a/ CuO+2HCl \to CuCl_2+H_2O\\ b/\\ n_{CuO}=0,125(mol)\\ \to n_{HCl}=0,125.2=0,25(mol)\\ m_{HCl}=0,25.36,5=9,125(g)\\ c/\\ n_{CuO}=n_{CuCl_2}=0,125(mol)\\ CM_{CuCl_2}=\frac{0,125}{0,5}=0,25M\)
a) \(CuO+2HCl\rightarrow CuCl2+H2O\)
b) Ta có: \(n_{CuO}=\dfrac{10}{80}=0,8\left(mol\right)\)
Theo PT: \(n_{HCl}=2nCuO=1,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,6.36,5=58,4\left(g\right)\)
c) \(n_{CuCl2}=n_{CuO}=0,8\left(mol\right)\)
\(V_{dd}=\)không đổi \(=500ml=0,5l\)
\(\Rightarrow C_{M\left(CuCl2\right)}=\dfrac{0,8}{0,5}=1,6\left(M\right)\)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Ta có: \(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,25\left(mol\right)\\n_{CuCl_2}=0,125\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,25\cdot36,5=9,125\left(g\right)\\C_{M_{CuCl_2}}=\dfrac{0,125}{0,5}=0,25\left(M\right)\end{matrix}\right.\)
a) \(n_{NaOH}=\dfrac{60.11,2\%}{40}=0,168\left(mol\right)\)
PTHH: \(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
0,168---->0,084----->0,084
b) \(m_{kt}=m_{Mg\left(OH\right)_2}=0,084.58=4,872\left(g\right)\)
c) \(C\%_{MgCl_2}=\dfrac{0,084.95}{190}.100\%=4,2\%\)
\(H_2SO_4+2NaOH->Na_2SO_4+2H_2O\\ H_2SO_4+Na_2CO_3->Na_2SO_4+CO_2+H_2O\\ n_{Na_2CO_3}=0,1mol=n_{H_2SO_4dư}\\ n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,15.2=0,15mol\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,15+0,1}{0,25}=1\left(M\right)\\ m_{Na_2SO_4}=142\left(0,15+0,1\right)=35,5g\)
500ml = 0,5l
\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
a) Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05 0,05
b) \(n_{Fe}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
⇒ \(m_{Fe}=0,05.56=2,8\left(g\right)\)
c) \(n_{H2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
d) \(n_{FeCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
⇒ \(m_{FeCl2}=0,05.127=6,35\left(g\right)\)
Chúc bạn học tốt
\(n_{SO_3}=\dfrac{8}{80}=0.1\left(mol\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(0.1......................0.1\)
\(C_{M_{H_2SO_4}}=\dfrac{0.1}{0.5}=0.2\left(M\right)\)
\(n_{CuO}=\dfrac{10}{80}=0.125\left(mol\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(0.1...........0.1\)
\(m_{CuO\left(dư\right)}=\left(0.125-0.1\right)\cdot80=2\left(g\right)\)