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\(A.HCl+NaOH\rightarrow NaCl+H_2O\\ B.n_{NaOH}=\dfrac{200.4}{100.40}=0,2mol\\ \Rightarrow\dfrac{1}{1}>\dfrac{0,4}{1}\Rightarrow HCl.dư\\ n_{NaCl}=n_{HCl,pư}=n_{NaOH}=0,2mol\\ m_{NaCl}=0,2.58,5=11,7g\\ m_{HCl,dư}=\left(1-0,2\right).36,5=29,2g\\ C_{\%NaCl}=\dfrac{11,7}{1.36,5+200}\cdot100=4,95\%\)
\(n_{NaOH}=\dfrac{150.20\%}{40}=0,75\left(mol\right)\\ n_{HCl}=\dfrac{250.7,3\%}{36,5}=0,5\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ Vì:\dfrac{0,75}{1}>\dfrac{0,5}{1}\Rightarrow NaOHdư\\ \Rightarrow n_{NaOH\left(p.ứ\right)}=n_{NaCl}=n_{HCl}=0,5\left(mol\right)\\ n_{NaOH\left(dư\right)}=0,75-0,5=0,25\left(mol\right)\\ C\%_{ddNaCl}=\dfrac{58,5.0,5}{150+250}.100=7,3125\%\\ C\%_{ddNaOH\left(dư\right)}=\dfrac{0,25.40}{150+250}.100=2,5\%\)
a, \(HCl+NaOH\rightarrow NaCl+H_2O\)
b, \(n_{HCl}=0,3.2=0,6\left(mol\right)\)
Theo PT: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{24}{20\%}=120\left(g\right)\)
c, \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{ZnCl_2}=0,3\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,3\cdot136=40,8\left(g\right)\\m_{ddHCl}=\dfrac{0,6\cdot36,5}{20\%}=109,5\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{19,5+109,5-0,6}\cdot100\%\approx37,78\%\)
a) NaOH + HCl --> NaCl + H2O
KOH + HCl --> KCl + H2O
b) Gọi số mol của NaOH, KOH là a, b (mol)
=> 40a + 56b = 3,04
Có nNaOH = nNaCl = a (mol)
=> mNaCl = 58,5a (g)
nKOH = nKCl = b (mol)
=> mKCl = 74,5b (g)
=> 58,5a + 74,5b = 4,15
=> a = 0,02; b = 0,04
\(\left\{{}\begin{matrix}m_{NaOH}=0,02.40=0,8\left(g\right)\\m_{KOH}=0,04.56=2,24\left(g\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}m_{NaCl}=0,02.58,5=1,17\left(g\right)\\m_{KCl}=0,04.74,5=2,98\left(g\right)\end{matrix}\right.\)
c)
PTHH: NaCl + AgNO3 --> NaNO3 + AgCl
0,02------------------------>0,02
KCl + AgNO3 --> KNO3 + AgCl
0,04--------------------->0,04
=> \(m_{AgCl}=\left(0,02+0,04\right).143,5=8,61\left(g\right)\)
\(a,NaOH+HCl\rightarrow NaCl+H_2O\\ KOH+HCl\rightarrow KCl+H_2O\\ b,Đặt:n_{NaOH}=w\left(mol\right);n_{KOH}=e\left(mol\right)\left(w,e>0\right)\\ \Rightarrow\left\{{}\begin{matrix}40w+56e=3,04\\58,5w+74,5e=4,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}w=0,02\left(mol\right)\\e=0,04\left(mol\right)\end{matrix}\right.\\ \Rightarrow m_{NaOH}=40w=0,8\left(g\right);m_{KOH}=56e=2,24\left(g\right)\\ c,NaCl+AgNO_3\rightarrow NaNO_3+AgCl\downarrow\\ KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\\ n_{AgCl\downarrow}=n_{NaCl}+n_{KCl}=w+e=0,06\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,06=8,61\left(g\right)\)
\(n_{Na_2CO_3}=0,1.1=0,1\left(mol\right)\)
a. \(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)
0,1 0,1 0,1 0,2
b. \(m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\)
c. \(C\%_{Ba\left(OH\right)_2}=\dfrac{0,1.171.100}{200}=8,55\%\)
d. \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\)
0,1 0,2
=> \(a=m_{dd.HCl}=\dfrac{0,2.36,5.100}{30}=\dfrac{73}{3}\left(g\right)\)
200ml = 0,2l
\(n_{HCl}=1.0,2=0,2\left(mol\right)\)
a) Pt : \(CaO+2HCl\rightarrow CaCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
b) \(n_{CaO}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CaO}=0,1.40=4\left(g\right)\)
c) \(n_{CaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CaCl2}=0,1.111=11,1\left(g\right)\)
d) \(C_{M_{CaCl2}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
Chúc bạn học tốt
Trả lời:
mk chx hok wa lớp 9 nên ko giúp đc, thông cảm
HT^^
\(NaOH+HCl->NaCl+H_2O\)
a, \(m_{HCl}=\frac{C\%.m_{\text{dd}HCl}}{100\%}=\frac{7,3\%.200}{100\%}=14.6g\)
\(n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{14.6}{36.5}=0.4\left(mol\right)\)
Theo PTHH ta có:\(n_{HCl}=n_{NaOH}=0.4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4.40=16g\)
\(\Rightarrow m_{\text{dd}NaOH}=\frac{m_{NaOH}.100\%}{C\%}=\frac{16.100\%}{10\%}=160g\)
b, Ta có \(\frac{C\%_{\text{dd}NaOH}-C\%_{\text{dd}mu\text{ối}}}{C\%_{\text{dd}mu\text{ối}}-C\%_{\text{dd}HCl}}=\frac{m_{\text{dd}HCl}}{m_{\text{dd}NaOH}}\)
\(\Leftrightarrow\frac{10\%-C\%}{C\%-7,3\%}=\frac{200}{160}=\frac{5}{4}\)\(\Rightarrow4\left(10\%-C\%\right)=5\left(C\%-7.3\%\right)\Leftrightarrow40\%-4C\%=5C\%-36.5\%\)
\(\Leftrightarrow9C\%=76.5\%\Leftrightarrow C\%=8,5\%\)
a)
$NaOH + HCl \to NaCl + H_2O$
b)
Theo PTHH :
$n_{HCl} = n_{NaOH} = \dfrac{80.20\%}{40}= 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$
c)
$n_{NaCl} = 0,4(mol) \Rightarrow m_{NaOH} = 0,4.58,5 = 23,4(gam)$