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a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b+c) Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,3\left(mol\right)\\n_{HCl}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,6\cdot36,5=21,9\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)
d) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) HCl còn dư, NaOH p/ứ hết
\(\Rightarrow\) Dung dịch sau p/ứ làm quỳ tím hóa đỏ
Theo PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=0,5\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,5\cdot58,5=29,25\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddHCl}+m_{NaOH}=\dfrac{0,6\cdot36,5}{5\%}+20=458\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{29,25}{458}\cdot100\%\approx6,39\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{458}\cdot100\%\approx0,8\%\end{matrix}\right.\)
a. PTHH: \(CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
b. Đổi 100ml = 0,1 lít
Ta có: \(n_{Cu\left(OH\right)_2}=\dfrac{9,8}{98}=0,1\left(mol\right)\)
Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)
=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)
c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)
=> \(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)
Ta có: \(n_{CuSO_4}=0,3\left(mol\right)\)
a, PT: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
______0,2____0,3_________________0,3 (mol)
b, \(m_{Al}=0,2.27=5,4\left(g\right)\)
c, \(m_{Cu}=0,3.64=19,2\left(g\right)\)
Bạn tham khảo nhé!
\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+H_2O\)
\(a\) \(2a\)
\(H_2SO_4+2KOH\rightarrow K_2SO_4+H_2O\)
\(b\) \(2b\)
Sau pư (1) đổi màu quỳ tìm \(\Rightarrow H_2SO_4\) dư \(n_{KON}=0,02.0,5=0,01\left(mol\right)\)
\(\rightarrow n_{H_2SO_4dư}=\frac{1}{2}n_{KOU}=5.10^{-3}\left(MOL\right)\)
\(\rightarrow n_{H_2O_4\text{ban đầu }}=0,05.1=0,05\left(mol\right)\)
\(\rightarrow n_{H_2SO_4\left(\text{pư 1 }\right)}=0,05-5.10^{-3}=0,045\left(mol\right)\)
\(\rightarrow n_{NaOH}=0,045.2=0,09\left(mol\right)\)
\(\rightarrow CM_{NaOH}=\frac{0,09}{0,05}=1,8\left(M\right)\)
\(PTHH:Fe+CuSO_4\)→\(FeSO_4+Cu\)
\(+n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
Theo PTHH ta có:
\(+n_{Cu}=n_{Fe}=0,3\left(mol\right)\)
\(+m_{Cu}=0,3.64=19,2\left(gam\right)\)
\(+n_{CuSO_4}=n_{Fe}=0,3\left(mol\right)\)
\(+V_{CuSO_4}=0,3.0,5=0,15\left(lit\right)\)
d)
PTHH: \(2NaOH+CuSO_4\) →\(Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
Cu(OH)2 làm quỳ tím chuyển màu xanh vì là bazo.