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\(1,PTHH:2NaOH+CuSO_4\to Na_2SO_4+Cu(OH)_2\downarrow\\ 2,n_{NaOH}=\dfrac{10}{40}=0,25(mol)\\ m_{CuSO_4}=\dfrac{160.20\%}{100\%}=32(g)\\ \Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2(mol)\)
Vì \(\dfrac{n_{NaOH}}{2}<\dfrac{n_{CuSO_4}}{1}\) nên \(CuSO_4\) dư
\(\Rightarrow n_{Cu(OH)_2}=0,125(mol)\\ \Rightarrow m_{Cu(OH)_2}=0,125.98=12,25(g)\\ 3,n_{Na_2SO_4}=0,125(mol)\\ \Rightarrow m_{CT_{Na_2SO_4}}=0,125.142=17,75(g)\\ m_{dd_{Na_2SO_4}}=10+160-12,25=157,75(g)\\ \Rightarrow C{\%}_{Na_2SO_4}=\dfrac{17,75}{157,75}.100\% \approx 11,25\%\)
Bài 1 :
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.1..................................0.1\)
\(m_{hh}=x=0.1\cdot56+4.4=10\left(g\right)\)
Bài 2 :
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0.1.......0.1..........0.1.............0.1\)
\(m_{Fe_2O_3}=7.2-0.1\cdot56=1.6\)
\(n_{Fe_2O_3}=\dfrac{7.2-0.1\cdot56}{160}=0.01\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
\(0.01...........0.03..............0.01\)
\(c.\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.1+0.03}{1}=0.13\left(l\right)\)
\(d.\)
\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.13}=\dfrac{10}{13}\left(M\right)\)
\(C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{0.03}{0.13}=\dfrac{3}{13}\left(M\right)\)
\(\begin{cases} m_{NaOH}=\dfrac{150.20\%}{100\%}=30(g)\\ m_{MgCl_2}=\dfrac{80.59,375\%}{100\%}=47,5(g) \end{cases} \Rightarrow \begin{cases} n_{NaOH}=\dfrac{30}{40}=0,75(mol)\\ n_{MgCl_2}=\dfrac{47,5}{95}=0,5(mol) \end{cases}\\ PTHH:2NaOH+MgCl_2\to Mg(OH)_2\downarrow+2NaCl\\ \text {Vì }\dfrac{n_{NaOH}}{2}<\dfrac{n_{MgCl_2}}{1} \text {nên }MgCl_2 \text { dư}\\ a,n_{Mg(OH)_2}=\dfrac{1}{2}n_{NaOH}=0,375(mol)\\ \Rightarrow m_{Mg(OH)_2}=0,375.58=21,75(g)\\ b,n_{NaCl}=m_{NaOH}=0,75(mol)\\ \Rightarrow m_{CT_{NaCl}}=0,75.58,5=43,875(g)\\ m_{dd_{NaCl}}=150+80-21,75=208,25(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{43,875}{208,25}.100\%\approx 21,07\%\)
Câu 1 :
Natri tan, lăn tròn trên mặt nước, xuất hiện khí không màu
$2Na + 2HCl \to 2NaCl + H_2$
Câu 2 :
a) $CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
b) $n_{CO_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$\Rightarrow n_{HCl} = 2n_{CO_2} = 0,15.2 =0,3(mol)$
$C_{M_{HCl}} = \dfrac{0,3}{0,2} = 1,5M$
c) $n_{CaCO_3} = n_{CO_2} = 0,15(mol)$
$\Rightarrow m_{NaCl} = 21 - 0,15.100 = 6\ gam$
\(n_{CuO}=\dfrac{1,6}{80}=0,02mol\\ n_{H_2SO_4}=\dfrac{100.20}{100.98}=\dfrac{10}{49}mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ \Rightarrow\dfrac{0,02}{1}< \dfrac{10:49}{1}\Rightarrow H_2SO_4.dư\\ n_{CuO}=n_{CuSO_4}=n_{H_2SO_4,pư}=0,02mol\\ C_{\%CuSO_4}=\dfrac{0,02.160}{1,6+100}\cdot100=3,15\%\\ C_{\%H_2SO_4}=\dfrac{\left(10:49-0,02\right)98}{1,6+100}\cdot100=17,76\%\%\)
Ta có: \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)
\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)
PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{CuSO_4}=n_{H_2SO_4\left(pư\right)}=n_{CuO}=0,02\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=\dfrac{10}{49}-0,02=\dfrac{451}{2450}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,02.160}{1,6+100}.100\%\approx3,15\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\dfrac{451}{2450}.98}{1,6+100}.100\%\approx17,76\%\end{matrix}\right.\)
\(a.n_{CuSO_4}=\dfrac{320.12,5}{100}:160=0,25mol\\ n_{NaOH}=\dfrac{150.20}{100}:40=0,75mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\\ \Rightarrow\dfrac{0,25}{1}< \dfrac{0,75}{2}\Rightarrow NaOH.dư\)
Đầu tiên màu xanh lam của \(CuSO_4\) nhạt dần rồi chuyển thành dung dịch không màu(NaOH dư) sau đó xuất hiện kết tủa màu xanh lam.
\(b.n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,25mol\\ n_{NaOH\left(dư\right)}=0,75-0,25.2=0,25mol\\ m_{Cu\left(OH\right)_2}=0,25.98=24,4g\\ m_{NaOH\left(dư\right)}=0,25.40=10g\\ c.m_{dd}=320+150=470g\\ C_{\%Cu\left(OH\right)_2}=\dfrac{24,4}{470}\cdot100=5,2\%\\ C_{\%NaOH\left(dư\right)}=\dfrac{10}{470}\cdot100=2,1\%\)
em không trừ kết tủa à em
⇒em nhầm ạ
sửa câu c
\(m_{dd}=320+150-24,4=445,6g\\ C_{\%NaOH}=\dfrac{10}{445,6}\cdot100=2,24\%\)