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\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\\ n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ \Rightarrow n_{KOH}=2n_{CuCl_2}=0,4\left(mol\right)\\ \Rightarrow C\%_{KOH}=\dfrac{0,4}{0,2}=2M\\ b,PTHH:Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\)
\(a.n_{CuCl_2}=\dfrac{13,5.10\%}{100\%.135}=0,01mol\\ CuCl_2+Ca\left(OH\right)_2\rightarrow Cu\left(OH\right)_2+CaCl_2\\ n_{Ca\left(OH\right)_2}=n_{Cu\left(OH\right)_2}=n_{CaCl_2}=0,01mol\\ m_{ddCa\left(OH\right)_2}=\dfrac{0,01.74}{25\%}\cdot100\%=2,96g\\ b.m_{Cu\left(OH\right)_2}=0,01.98=0,98g\\ c.m_{dd}=13,5+2,96-0,98=15,48g\\ C_{\%CaCl_2}=\dfrac{0,01.111}{15,48}\cdot100\%=7,17\%\\ d.Cu\left(OH\right)_2\xrightarrow[]{t^0}CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=0,01mol\\ m_{CuO}=0,01.80=0,8g\)
\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
a) \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\left(1\right)\)
\(Cu\left(OH\right)_2\xrightarrow[t^o]{}CuO+H_2O\left(2\right)\)
b) \(Pt\left(1\right):n_{Cu\left(OH\right)2}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(Pt\left(2\right):n_{Cu\left(OH\right)2}=n_{CuO}=0,25\left(mol\right)\Rightarrow m_{Cu}=0,25.64=16\left(g\right)\)
c) Pt(1) : \(n_{NaOH}=n_{NaCl}=0,5\left(mol\right)\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)
a, \(n_{CuCl_2}=\dfrac{48,5}{135}=\dfrac{97}{270}\left(mol\right)\)
\(n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\)
PT: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)
Xét tỉ lệ: \(\dfrac{\dfrac{97}{270}}{1}>\dfrac{0,6}{2}\), ta được CuCl2 dư.
Theo PT: \(n_{CuCl_2\left(pư\right)}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,3\left(mol\right)\)
\(\Rightarrow m_{CuCl_2\left(dư\right)}=48,5-0,3.135=8\left(g\right)\)
b, \(m_{Cu\left(OH\right)_2}=0,3.98=29,4\left(g\right)\)
\(m_{CuCl_2}=100\times25\%=25\left(g\right)\)
\(\Rightarrow n_{CuCl_2}=\dfrac{25}{135}=\dfrac{5}{27}\left(mol\right)\)
\(m_{KOH}=140\times20\%=28\left(g\right)\)
\(\Rightarrow n_{KOH}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: CuCl2 + 2KOH → 2KCl + Cu(OH)2↓
Ban đầu: \(\dfrac{5}{27}\) ...........0,5........................................... (mol)
Phản ứng: \(\dfrac{5}{27}\) ...........\(\dfrac{10}{27}\) ......................................... (mol)
Sauphảnứng: 0..............\(\dfrac{7}{54}\) ..→ \(\dfrac{10}{27}\) ...........\(\dfrac{5}{27}\) ......... (mol)
a) \(m_{Cu\left(OH\right)_2}=\dfrac{5}{27}\times98=18,15\left(g\right)\)
b) \(m_{dd}saupư=100+140-18,15=221,85\left(g\right)\)
\(m_{KOH}dư=\dfrac{7}{54}\times56=7,26\left(g\right)\)
\(\Rightarrow C\%_{KOH}dư=\dfrac{7,26}{221,85}\times100\%=3,27\%\)