Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)
c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)
\(n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2\left(mol\right)\)
PTHH :
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
0,2 0,4 0,2 0,2
\(m_{NaOH}=0,4.40=16\left(g\right)\)
\(m_{ddNaOH}=\dfrac{16.100}{10}=160\left(g\right)\)
\(c,m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
\(m_{ddNa_2SO_4}=200+160-\left(0,2.98\right)=340,4\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{28,4}{240,4}.100\%\approx8,34\%\)
\(d,PTHH:\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
0,2 0,2
\(m_{CuO}=0,2.80=16\left(g\right)\)
a, \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b, \(m_{CuSO_4}=200.16\%=32\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{NaOH}=2n_{CuSO_4}=0,4\left(mol\right)\Rightarrow m_{ddNaOH}=\dfrac{0,4.40}{10\%}=160\left(g\right)\)
c, \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,2.142}{200+160-0,2.98}.100\%\approx8,34\%\)
d, \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)
a, \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
b, \(m_{CuSO_4}=250.16\%=40\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{40}{160}=0,25\left(mol\right)\)
Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,25\left(mol\right)\)
\(\Rightarrow a=m_{CuO}=0,25.80=20\left(g\right)\)
c, Ta có: m dd sau pư = m dd NaOH + m dd CuSO4 - mCu(OH)2 = 200 + 250 - 0,25.98 = 425,5 (g)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)
\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
\(CuCl_2+2NaOH--->Cu\left(OH\right)_2\downarrow+2NaCl\left(1\right)\)
0,1_______0,3x______________0,1________0,2
\(Cu\left(OH\right)_2--to->CuO+CO_2\uparrow\left(2\right)\)
0,1_______________0,1
\(n_{CuCl_2}=0,2.0,5=0,1\left(mol\right)\)
\(n_{NaOH}=0,3x\left(mol\right)\)
b) =>\(0,3x=0,1.2=>x=0,67\left(M\right)\)
=> \(m=0,1.80=8\left(g\right)\)
c) => \(C_{M_{NaCl}}=\frac{0,2}{0,5}=0,4\left(M\right)\)
a.CuCl2 + 2NaOH -> Cu(OH)2 + 2NaCl
0.15 0.3 0.15 0.3
Cu(OH)2 -> CuO + H2O
0.15 0.15
nNaOH = 0.3 mol
\(CM_{CuCl2}=\dfrac{0.15}{2}=0.075M\)
b.Vdd sau phản ứng = 0.2 + 0.15 = 0.35l
\(CM_{NaCl}=\dfrac{0.3}{0.35}=0.86M\)
c.mCuO = \(0.15\times80=12g\)
\(m_{NaOH}=\dfrac{100\cdot10\%}{100\%}=10g\) \(\Rightarrow n_{NaOH}=0,25mol\)
\(ZnCl_2+2NaOH\rightarrow Zn\left(OH\right)_2+2NaCl\)
0,025 0,05 0,025
\(Zn\left(OH\right)_2\underrightarrow{t^o}ZnO+H_2O\)
0,025 0,025
\(m=m_{ZnO}=0,025\cdot\left(65+16\right)=2,025g\)
\(C_{M_{ZnCl_2}}=\dfrac{0,025}{\dfrac{500}{1000}}=0,05M\)
tại sao chia cho 1000 thế ạ