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\(n_{KOH}=\dfrac{100.14}{100.56}=0,25(mol)\\ 2KOH+CuCl_2\to Cu(OH)_2\downarrow+2KCl\\ \Rightarrow n_{CuCl_2}=n_{Cu(OH)_2}=0,125(mol);n_{KCl}=0,25(mol)\\ a,m_{CuCl_2}=0,125.135=16,875(g)\\ b,m_{Cu(OH)_2}=0,125.98=12,25(g)\\ c,C\%_{KCl}=\dfrac{0,25.74,5}{100+16,875-12,25}.100\%=17,8\%\\ d,Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=0,125(mol)\\ \Rightarrow m_{CuO}=0,125.80=10(g)\)
a/ PTHH : \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\)
b/ \(n_{CO_2}=\frac{22,4}{22,4}=1\left(mol\right)\)
Từ PTHH suy ra \(n_{Ba\left(OH\right)_2}=n_{CO_2}=1\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\frac{n_{Ba\left(OH\right)_2}}{V_{Ba\left(OH\right)_2}}=\frac{1}{\frac{200}{1000}}=5M\)
c/ \(n_{BaCO_3}=n_{CO_2}=1\left(mol\right)\Rightarrow m_{BaCO_3}=1\times197=197\left(g\right)\)
\(n_{MgCO3}=\dfrac{8,4}{84}=0,1\left(mol\right)\)
a) Pt : \(2CH_3COOH+MgCO_3\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O|\)
2 1 1 1 1
0,2 0,1 0,1 0,1
b) \(n_{CH3COOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{CH3COOH}=0,2.60=12\left(g\right)\)
\(C_{ddCH3COOH}=\dfrac{12.100}{200}=6\)0/0
\(n_{\left(CH3COO\right)2Mg}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{\left(CH3COO\right)2Mg}=0,1.142=14,2\left(g\right)\)
\(m_{ddspu}=8,4+200-\left(0,1.44\right)=204\left(g\right)\)
\(C_{dd\left(CH3COO\right)2Mg}=\dfrac{14,2.100}{204}=6,96\)0/0
Chúc bạn học tốt
Câu 1
a)CuSO4+2NaOH---->Cu(OH)2+Na2SO4
Ta có
n CuSO4=0,4.2=0,8(mol)
Theo pthh
n Cu(OH)2=n CuSO4=0,8(mol0
m Cu(OH)2=0,8.98=78,4(g)
b)Cu(OH)2---->CuO+H2O
Theopthh2
n CuO=n Cu(OH)2=0,8(MOL)
m=m CuO=0,8.80=64(g)
Bài 2
2A+Cl2--->2ACl
Ta có
n A=\(\frac{9,2}{A}\)
n ACl=\(\frac{23,4}{A+35,5}\)
Theo pthh
n A=n ACl---->\(\frac{9,2}{A}=\frac{23,4}{A+35,5}\)
-->23,4A=9,2A+326,6
---->14,2A=326,6
>A=23
Vậy A là Na
Câu 3
Mg+2HCl---.MgCl2+H2
Fe+2HCl---->FeCl2+H2
Zn+2HCl-->ZnCl2+H2
n H2=8,96/22,4=0,4(mol)
Theo pthh 1,2,3
n HCl=2n H2=0,8(mol)
m HCl=0,8.36,5=29,2(g)
Áp dụng ĐLBTKL ta có
m Muối=m KL+m HCl-m H2
=16,9+29,2-0,845,3(g)
PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,2\cdot2=0,4\left(mol\right)\\n_{NaOH}=0,2\cdot2=0,4\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) \(\Rightarrow\) CuCl2 còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,4\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}=n_{CuCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2\cdot80=16\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,2+0,2}=1\left(M\right)\\C_{M_{CuCl_2}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\\\end{matrix}\right.\)
PT: \(Cu\left(NO_3\right)_2+2NaOH\rightarrow2NaNO_3+Cu\left(OH\right)_{2\downarrow}\)
a, Ta có: \(m_{Cu\left(NO_3\right)_2}=\frac{50.5\%}{100\%}=2,5\left(g\right)\)
\(\Rightarrow n_{Cu\left(NO_3\right)_2}=\frac{2,5}{188}=\)
Tới đây tính ra số mol Cu(NO3)3 lẻ quá, không biết đề có nhầm lẫn ở đâu không bạn nhỉ?
PTHH: \(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaNO_3\)
a) Ta có: \(m_{Cu\left(NO_3\right)_2}=50\cdot5\%=2,5\left(g\right)\) \(\Rightarrow n_{Cu\left(NO_3\right)_2}=\frac{2,5}{188}=\frac{5}{376}\left(mol\right)\)
\(\Rightarrow n_{NaOH}=\frac{5}{188}\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\frac{\frac{5}{188}}{0,06}\approx0,44\left(M\right)\\m_{NaOH}=\frac{5}{188}\cdot40=\frac{50}{47}\left(g\right)\end{matrix}\right.\)
Theo đề bài: \(m_{ddNaOH}=60\cdot1,2=72\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{\frac{50}{47}}{72}\cdot100\approx1,48\%\)
b) Theo PTHH: \(n_{Cu\left(NO_3\right)_2}=n_{Cu\left(OH\right)_2}=\frac{5}{376}\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=\frac{5}{376}\cdot98\approx1,3\left(g\right)\)
c) Theo PTHH: \(n_{NaOH}=n_{NaNO_3}=\frac{5}{188}\left(mol\right)\) \(\Rightarrow m_{NaNO_3}=\frac{5}{188}\cdot85\approx2,26\left(g\right)\)
Mặt khác: \(m_{dd}=m_{Cu\left(NO_3\right)_2}+m_{NaOH}-m_{Cu\left(OH\right)_2}=50+72-1,3=120,7\left(g\right)\)
\(\Rightarrow C\%_{NaNO_3}=\frac{2,26}{120,7}\cdot100\approx1,87\%\)
câu 1
cho 2dd trên td vs NaOH dư
có tủa => CuSO4
CuSO4 + 2NaOH => Na2SO4 + Cu(OH)2
ko hiện tượng => Na2SO4
a. PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
b. Có \(n_{Mg}=\frac{3,6}{24}=0,15mol\)
\(140ml=0,14l\)
\(n_{H_2SO_4}=0,14.1,2=0,168mol\)
Lập tỉ lệ \(\frac{n_{Mg}}{1}< \frac{n_{H_2SO_4}}{1}\)
Vậy Mg đủ, \(H_2SO_4\) dư
Theo phương trình \(n_{H_2SO_4}=n_{Mg}=0,15mol\)
\(\rightarrow n_{H_2SO_4\left(\text{(dư)}\right)}=0,168-0,15=0,018mol\)
\(\rightarrow m_{H_2SO_4\left(\text{(dư)}\right)}n.M=0,018.98=1,764g\)
c. MgSO\(_4\) là muối
Theo phương trình \(n_{MgSO_4}=n_{Mg}=0,15mol\)
\(\rightarrow m_{\text{muối}}=m_{MgSO_4}=n.M=0,15.120=18g\)
d. \(H_2\) là khí
Theo phương trình \(n_{H_2}=n_{Mg}=0,15mol\)
\(\rightarrow V_{H_2\left(ĐKTC\right)}=n.22,4=0,15.22,4=3,36l\)
a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
b, \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
Theo PT: \(n_{CuCl_2}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)
\(\Rightarrow m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)
c, \(C_{M_{CuCl_2}}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)
\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
PTHH:
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
0,125 0,25 0,125 0,25
\(m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)
\(C_{M\left(CuCl_2\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)
\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
a) PTHH : \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
0,125<-----0,25---------------------->0,125
b) \(m_{Kt}=m_{Cu\left(OH\right)2}=0,125.98=12,25\left(g\right)\)
c) \(C_{MCuCl2}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)
Số mol đặt dưới PT bị nhầm rồi em nhé.