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2)
Zn+ H2SO4 ---> ZnSO4+ H2↑
0.1 0.1
nH2= \(\dfrac{2,24}{22,4}\)0.1 mol
mZn= 0.1x65=6.5 g
mCu=10.5-6,5=4 g
%mZn=\(\dfrac{6,5}{10,5}\).100%=61.9%
%mCu=\(\dfrac{4}{10,5}\)x100%=38.1%
\(n_{CO_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\\ n_{NaOH}=n_{OH}=1.0,38=0,38\left(mol\right)\\ n_{Ba\left(OH\right)_2}=1.0,1=0,1\left(mol\right)\)
\(TL:\frac{n_{OH}}{n_{CO_2}}=\frac{0,38}{0,3}=1,26\)
→ Tạo ra hh 2 muối
\(PTHH:CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ PTHH:CO_2+NaOH\rightarrow NaHCO_3\)
\(n_{Na_2CO_3}=x;n_{NaHCO_3}=y\)
\(\Rightarrow hpt:\left\{{}\begin{matrix}2x+y=0,38\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,08\\y=0,22\end{matrix}\right.\)
\(PTHH:Ba\left(OH\right)_2+Na_2CO_3\rightarrow BaCO_3+2NaOH\\ PTHH:Ba\left(OH\right)_2+2NaHCO_3\rightarrow BaCO_3+Na_2CO_3+2H_2O\)
\(\Rightarrow\left\{{}\begin{matrix}n_{BaCO_3\left(1\right)}=0,08\left(mol\right)\\n_{BaCO_3\left(2\right)}=0,11\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{kt}=197.\left(0,08+0,11\right)=37,43\left(g\right)\)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(n_{NaOH}=0.2\left(mol\right)\)
\(n_{FeCl_3}=0.1\cdot1=0.1\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\)
\(3....................1\)
\(0.2.............0.1\)
\(LTL:\dfrac{0.2}{3}< \dfrac{0.1}{1}\Rightarrow FeCl_3dư\)
\(m_{Fe\left(OH\right)_3}=\dfrac{0.2}{3}\cdot107=7.13\left(g\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(n_{Na}=n_{NaOH}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_{Fe\left(OH\right)_3}=0,1.1=0,1\left(mol\right)\)
Lập tỉ lệ : \(\dfrac{0,2}{3}< \dfrac{0,1}{1}\) => FeCl3 dư, NaOH hết
\(n_{Fe\left(OH\right)_3}=\dfrac{1}{3}n_{NaOH}=\dfrac{0,2}{3}=\dfrac{1}{15}\left(mol\right)\)
=> \(m_{Fe\left(OH\right)_3}=\dfrac{1}{15}.107=7,13\left(g\right)\)
PTHH: \(Na_2SO_4+CaCl_2\rightarrow2NaCl+CaSO_4\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,1\cdot0,5=0,05\left(mol\right)\\n_{CaCl_2}=0,1\cdot0,4=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Na2SO4 dư
\(\Rightarrow\left\{{}\begin{matrix}n_{CaSO_4}=0,04\left(mol\right)\\n_{NaCl}=0,08\left(mol\right)\\n_{Na_2SO_4\left(dư\right)}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaSO_4}=0,04\cdot136=5,44\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,08}{0,1+0,1}=0,4\left(M\right)\\C_{M_{Na_2SO_4\left(dư\right)}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=1\cdot0,1=0,1\)
\(Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\uparrow\)
0,05 ← 0,1 → 0,05
\(\Rightarrow V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{HCl}}\Rightarrow n_{HCl}=C_{M_{HCl}}.V_{HCl}=1.\left(100:1000\right)=0,1\left(mol\right)\)
\(PTHH:Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)
Theo phương trình Fe dư
\(PTHH:Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)
tỉ lệ :1 2 1 1(mol)
Số mol :0,05 0,05 0,05 0,05(mol)
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
lẹ nhasai đề rồi bạn, số mol kết tủa không thể lớn hơn số mol Ca(OH)2
600ml = 0,6l
600ml = 0,6l
\(n_{H2SO4}=1.0,6=0,6\left(mol\right)\)
\(n_{BaCl2}=0,5.0,6=0,3\left(mol\right)\)
Pt : \(H_2SO_4+BaCl_2\rightarrow2HCl+BaSO_4|\)
1 1 2 1
0,6 0,3 0,6 0,3
a) Lập tỉ số so sánh : \(\dfrac{0,6}{1}>\dfrac{0,3}{1}\)
⇒ H2SO4 dư , BaCl2 phản ứng hết
⇒ Tính toán dựa vào số mol của BaCl2
\(n_{BaSO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{BaSO4}=0,3.233=69,9\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,3.2}{1}=0,6\left(mol\right)\)
\(n_{H2SO4\left(dư\right)}=0,6-0,3=0,3\left(mol\right)\)
\(V_{ddspu}=0,6+0,6=1,2\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0,6}{1,2}=0,5\left(M\right)\)
\(C_{M_{H2SO4\left(dư\right)}}=\dfrac{0,3}{1,2}=0,25\left(M\right)\)
Chúc bạn học tốt
\(n_{Ba\left(OH\right)_2}=0,1.1=0,1\left(mol\right)\)
\(n_{H_2SO_4}=0,1.0,8=0,08\left(mol\right)\)
PTHH:
\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+H_2O\)
0,08 0,08 0,08 0,08
\(\dfrac{0,1}{1}>\dfrac{0,08}{1}\)--> Ba(OH)2 dư
\(m_{BaSO_4}=0,08.233=18,46\left(g\right)\)