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Sửa: \(14,7\%\)
\(n_{H_2SO_4}=\dfrac{200.14,7\%}{100\%.98}=0,3(mol)\\ a,PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ b,n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=0,2(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g)\\ c,n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{200+5,4-0,3.2}.100\%=16,7\%\\ c,m_{Al_2(SO_4)_3}=0,1.342=34,2(g)\)
\(n_{Al_2\left(SO_4\right)_3}=0,3.0,2=0,06\left(mol\right)\)
\(n_{Al\left(OH\right)_3}=\dfrac{1,56}{78}=0,02\left(mol\right)\)
TH1: NaOH vừa đủ:
\(\dfrac{NaOH}{0,02\left(mol\right)}+\dfrac{Al_2\left(SO_4\right)_3}{0,02\left(mol\right)}\rightarrow\dfrac{Al\left(OH\right)_3\downarrow}{0,02\left(mol\right)}+Na_2SO_4\)
\(\Rightarrow C_{M\left(NaOH\right)}=\dfrac{0,02}{0,5}=0,1\left(M\right)\)
TH2: NaOH dư.
\(\dfrac{NaOH}{0,06\left(mol\right)}+\dfrac{Al_2\left(SO_4\right)_3}{0,06\left(mol\right)}\rightarrow\dfrac{Al\left(OH\right)_3\downarrow}{0,06\left(mol\right)}+Na_2SO_4\)(1)
\(\dfrac{NaOH}{0,02\left(mol\right)}+\dfrac{Al\left(OH\right)_3}{0,02\left(mol\right)}\rightarrow NaAlO_2+2H_2O\)(2)
Từ (1) (2), Suy ra: \(n_{NaOH}=0,06+0,02=0,08\left(mol\right)\)
\(C_{M\left(NaOH\right)}=\dfrac{0,08}{0,5}=0,16\left(M\right)\)
a)mH2SO4=\(\dfrac{200.7,3\text{%}}{100\%}\)=14,6g
nHCl=\(\dfrac{14,6}{36,5}\)=0,4(mol)
PTHH:
NaOH+ HCl→ NaCl+ H2O
1 1 1 1
0,4 0,4 0,4 (mol)
⇒mNaOH=0,4.40=16(g)
Nồng độ % của dd NaOH cần dùng là:
C%NaOH=\(\dfrac{16}{200}\) .100%=8%
b)Ta có:mdd spứ=mdd trc pứ=400g
mNaCl=0,4.58,5=23,4g
Nồng độ % dd muối tạo thành sau pứ là:
C%dd NaCl=\(\dfrac{23,4}{400}\) .100%=5,85%
PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2.
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{200}.100\%=14,7\%\)
=> \(m_{H_2SO_4}=29,4\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.n_{H_2SO_4}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
=> \(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
Bài 1:
a, Hiện tượng: Có khí mùi hắc thoát ra.
b, Ta có: \(m_{H_2SO_4}=100.24,5\%=24,5\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)
PT: \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)
Theo PT: \(n_{Na_2SO_3}=n_{H_2SO_4}=0,25\left(mol\right)\)
\(\Rightarrow C\%_{Na_2SO_3}=\dfrac{0,25.126}{200}.100\%=15,75\%\)
c, Theo PT: \(n_{SO_2}=n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)
⇒ m dd sau pư = 200 + 100 - 0,25.64 = 284 (g)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,25.142}{284}.100\%=12,5\%\)
Bài 2:
a, Hiện tượng: Xuất hiện kết tủa trắng.
PT: \(BaCl_2+MgSO_4\rightarrow MgCl_2+BaSO_{4\downarrow}\)
b, Ta có: \(m_{BaCl_2}=200.20,8\%=41,6\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)
Theo PT: \(n_{MgCl_2}=n_{BaSO_4}=n_{MgSO_4}=n_{BaCl_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{ddMgSO_4}=\dfrac{0,2.120}{12\%}=200\left(g\right)\)
c, Ta có: m dd sau pư = 200 + 200 - 0,2.233 = 353,4 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95}{353,4}.100\%\approx5,38\%\)
PTHH:\(Na_2SO_3+CaCl_2\rightarrow2NaCl+CaSO_3\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_3}=\dfrac{265\cdot10\%}{126}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2}=\dfrac{500\cdot6,66\%}{111}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỷ số: \(\dfrac{53}{252}< \dfrac{0,3}{1}\) \(\Rightarrow\) CaCl2 còn dư, Na2SO3 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=\dfrac{53}{126}\left(mol\right)\\n_{CaSO_3}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=\dfrac{53}{126}\cdot58,5\approx24,61\left(g\right)\\m_{CaSO_3}=\dfrac{53}{252}\cdot120\approx25,24\left(g\right)\\m_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\cdot111\approx9,95\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNa_2SO_3}+m_{ddCaCl_2}-m_{CaSO_3}=739,76\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{24,61}{739,76}\cdot100\%\approx3,33\%\\C\%_{CaCl_2\left(dư\right)}=\dfrac{9,95}{739,76}\cdot100\%\approx1,35\%\end{matrix}\right.\)
a) \(NaOH+HCl\rightarrow NaCl+H_2O\)
0,1................0,3
LẬp tỉ lệ : \(\dfrac{0,1}{1}< \dfrac{0,3}{1}\)=> Sau pứ HCl dư
\(m_{NaCl}=0,1.58,5=5,85\left(g\right)\)
b) \(CM_{NaCl}=\dfrac{0,1}{0,2+0,3}=0,2M\)
\(CM_{HCl\left(dư\right)}=\dfrac{\left(0,3-0,1\right)}{0,2+0,3}=0,4M\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{200\cdot10\%}{342}=\dfrac{10}{171}\left(mol\right)\)
\(n_{NaOH}=\dfrac{500\cdot20\%}{40}=2.5\left(mol\right)\)
\(6NaOH+Al_2\left(SO_4\right)_3\rightarrow3Na_2SO_4+2Al\left(OH\right)_3\)
\(6..................1\)
\(2.5..................\dfrac{10}{171}\)
\(LTL:\dfrac{2.5}{6}>\dfrac{10}{171}\Rightarrow NaOHdư\)
\(n_{NaOH\left(dư\right)}=2.5-\dfrac{10}{171}\cdot6=2.15\left(mol\right)\)
\(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+H_2O\)
\(\dfrac{20}{171}........\dfrac{20}{171}.........\dfrac{20}{171}\)
\(n_{NaOH\left(cl\right)}=2.15-\dfrac{20}{171}\approx2\left(mol\right)\)
\(m_{\text{dung dịch sau phản ứng}}=200+500=700\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{\dfrac{10}{57}\cdot142}{700}\cdot100\%=3.55\%\)
\(C\%_{NaAlO_2}=\dfrac{\dfrac{20}{171}\cdot82}{700}\cdot100\%=1.37\%\)
$n_{Al_2(SO_4)_3} = \dfrac{200.10\%}{342} = \dfrac{10}{171}(mol)$
$n_{NaOH} = \dfrac{500.20\%}{40} = 2,5(mol)$
$Al_2(SO_4)_3 + 6NaOH \to 2Al(OH)_3 + 3Na_2SO_4$
Ta thấy :
\(\dfrac{n_{Al_2(SO_4)_3}}{1} > \dfrac{n_{NaOH}}{6}\) nên NaOH dư
Theo PTHH :
n NaOH pư = 6n Al2(SO4)3 = 20/57(mol)
n Na2SO4 = 3n Al2(SO4)3 = 10/57(mol)
n Al(OH)3 = 2n Al2(SO4)3 = 20/171(mol)
Sau pư :
m dd = 200 + 500 - 78.20/171 = 709,122(gam)
\(C\%_{NaOH} = \dfrac{(2,5 - \dfrac{20}{57})40}{709,122}.100\% = 12,12\%\\ C\%_{Na_2SO_4} = \dfrac{\dfrac{10}{57}.142}{709,122}.100\% = 3,51\%\)