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\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
______0,05------>0,15--------->0,05
=> mH2SO4 = 0,15.98 = 14,7(g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)
\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)
PTHH: Fe2(SO4)3 + 6NaOH --> 2Fe(OH)3\(\downarrow\) + 3Na2SO4
________0,05----------------------->0,1
=> mFe(OH)3 = 0,1.107=10,7(g)
\(a,n_{Na_2CO_3}=\dfrac{106.10}{100.106}=0,1mol\\ BaCl_2+Na_2CO_3\rightarrow BaCO_3+2NaCl\\ n_{BaCO_3}=n_{Na_2CO_3}=0,1mol\\ m_A=m_{BaCO_3}=0,1.197=19,7g\\ b,n_{NaCl}=0,1.2=0,2mol\\ C_{\%B}=C_{\%NaCl}=\dfrac{0,2.58,5}{100+106-19,7}\cdot100=6,28\%\\ c.BaCO_3\xrightarrow[]{t^0}BaO+CO_2\\ n_{CO_2}=n_{BaCO_3}=0,1mol\\ n_{Ca\left(OH\right)_2}=0,08.1=0,08mol\\ T=\dfrac{0,08}{0,1}=0,8\\ \Rightarrow0,5< T< 1\)
Pứ tạo 2 muối
\(n_{CaCO_3}=a,n_{Ca\left(HCO_3\right)_2}=b\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ 2CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(HCO_3\right)_2\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,08\\a+2b=0,1\end{matrix}\right.\\ \Rightarrow a=0,06;b=0,02\\ m_{muối}=0,06.100+0,02.162=9,24g\)
\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ ...0,2......0,4.......0,2........0,4\left(mol\right)\\ b,n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\ c,m_{KOH}=0,4\cdot56=22,4\left(g\right)\\ m_{dd_{KOH}}=\dfrac{22,4\cdot100\%}{20\%}=112\left(g\right)\\ m_{dd_{KCl}}=m_{CuCl_2}+m_{dd_{KOH}}-m_{Cu\left(OH\right)_2}=27+112-19,6=119,4\left(g\right)\)
\(d,C\%_{dd_{KCl}}=\dfrac{74,5\cdot0,4}{119,4}\cdot100\%\approx24,96\%\)
a)
$AgNO_3 + HCl \to AgCl + HNO_3$
Theo PTHH :
$n_{AgCl} = n_{HCl} = n_{AgNO_3} = \dfrac{340.10\%}{170} =0,2(mol)$
$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$
b)
$m_{AgCl} = 0,2.143,5 = 28,7(gam)$
c)
$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$
$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$
\(n_{BaCl_2}=\dfrac{31,2}{208}=0,15mol\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,15 0,15 0,15 0,3
a)\(m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\)
b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6}\cdot100=75\left(g\right)\)
c)\(m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)
\(m_{ddsau}=31,2+75-34,95=71,25\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{10,95}{71,25}\cdot100\%=15,37\%\)
\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,095 0,19 0,095 0,095
\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)
0,095 0,095
\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)
Câu 1:
PTHH: 2Al + 3H2SO4 ===> Al2(SO4)3 + 3H2
a)Vì Cu không phản ứng với H2SO4 loãng nên 6,72 lít khí là sản phẩm của Al tác dụng với H2SO4
=> nH2 = 6,72 / 22,4 = 0,2 (mol)
=> nAl = 0,2 (mol)
=> mAl = 0,2 x 27 = 5,4 gam
=> mCu = 10 - 5,4 = 4,6 gam
b) nH2SO4 = nH2 = 0,3 mol
=> mH2SO4 = 0,3 x 98 = 29,4 gam
=> Khối lượng dung dịch H2SO4 20% cần dùng là:
mdung dịch H2SO4 20% = \(\frac{29,4.100}{20}=147\left(gam\right)\)
nH2 = 6.72 : 22.4 = 0.3 mol
Cu không tác dụng với H2SO4
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
0.2 <- 0.3 <- 0.1 <- 0.3 ( mol )
mAl = 0.2 x 56 = 5.4 (g)
mCu = 10 - 5.4 = 4.6 (g )
mH2SO4 = 0.3 x 98 = 29.4 ( g)
mH2SO4 20% = ( 29.4 x100 ) : 20 = 147 (g)
a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$
b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)
Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
a, \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{0,6.36,5}{500}.100\%=4,38\%\)
b, \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
PT: \(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)
______0,2_______0,6______________0,2 (mol)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
\(m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)
a) \(n_{K2SO4}=\dfrac{17,4}{174}=0,1\left(mol\right)\)
PTHH : \(K_2SO_4+BaCl_2-->BaSO_4\downarrow+2KCl\)
Theo PTHH :nBaSO4 = nK2SO4 = 0,1 (mol)
=> mBaSO4 = 0,1. 233 = 23,3 (g)
b) Theo PTHH :
nKCl = 2nK2SO4 = 0,2 (mol)
nBaCl2 = nK2SO4 = 0,1 (mol)
=> mBaCl2 = 0,1.208 = 20,8 (g)
=> m(ddBaCl2) = 20,8 : 10.100 = 208 (g)
Áp dụng định luật bảo toàn khối lượng :
mK2SO4 + m(ddBaCl2) = mBaSO4 + m(ddKCl)
=> 17,4 + 208 = 23,3 + m(ddKCl)
=> m(ddKCl) = 202,1 (g)
=> \(C\%KCl=\dfrac{0,2.74,5}{202,1}\cdot100\%\approx7,37\%\)
\(\begin{array}{l} a,\\ n_{K_2SO_4}=\dfrac{17,4}{174}=0,1\ (mol)\\ PTHH:K_2SO_4+BaCl_2\to BaSO_4\downarrow+2KCl\\ Theo\ pt:\ n_{BaSO_4}=n_{K_2SO_4}=0,1\ (mol)\\ \Rightarrow m_{BaSO_4}=0,1\times 233=23,3\ (g)\\ b,\\ Theo\ pt:\ n_{BaCl_2}=n_{K_2SO_4}=0,1\ (mol)\\ \Rightarrow m_{\text{dd BaCl_2}}=\dfrac{0,1\times 208}{10\%}=208\ (g)\\ m_{\text{dd spư}}=m_{K_2SO_4}+m_{\text{dd BaCl_2}}-m_{BaSO_4}\\ \Rightarrow m_{\text{dd spư}}=17,4+208-23,3=202,1\ (g)\\ Theo\ pt:\ n_{KCl}=2n_{K_2SO_4}=0,2\ (mol)\\ \Rightarrow C\%_{\text{dd spư}}=C\%_{KCl}=\dfrac{0,2\times 74,5}{202,1}\times 100\%=7,37\%\end{array}\)