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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 \(\frac{C\%_{\text{dd}NaOH}-C\%_{\text{dd}mu\text{ối}}}{C\%_{\text{dd}mu\text{ối}}-C\%_{\text{dd}HCl}}=\frac{m_{\text{dd}HCl}}{m_{\text{dd}NaOH}}\)
\(\Leftrightarrow\frac{10\%-C\%}{C\%-7,3\%}=\frac{200}{160}=\frac{5}{4}\)\(\Rightarrow4\left(10\%-C\%\right)=5\left(C\%-7.3\%\right)\Leftrightarrow40\%-4C\%=5C\%-36.5\%\)
\(\Leftrightarrow9C\%=76.5\%\Leftrightarrow C\%=8,5\%\)
\(m_{ct}=\dfrac{9,8.150}{100}=14,7\left(g\right)\)
\(n_{H2SO4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
\(n_{MgO}=\dfrac{10}{40}=0,25\left(mol\right)\)
Pt : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O|\)
1 1 1 1
0,25 0,15 0,15
a) Lap ti so so sanh : \(\dfrac{0,25}{1}>\dfrac{0,15}{1}\)
⇒ MgO du , H2SO4 phan ung het
⇒ Tinh toan dua vao so mol cua H2SO4
\(n_{MgSO4}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{MgSO4}=0,15.120=18\left(g\right)\)
b) \(m_{ddspu}=150+10=160\left(g\right)\)
\(C_{MgSO4}=\dfrac{18.100}{160}=11,25\)0/0
Chuc ban hoc tot
\(n_{NaOH}=1.0,5=0,5(mol)\\ 2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,25(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,25.98}{9,8\%}=250(g)\)
a) 500ml = 0,5l
\(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
b) \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O|\)
2 1 1 2
0,3 0,15
\(n_{H2SO4}=\dfrac{0,3.1}{2}=0,15\left(mol\right)\)
\(m_{H2SO4}=0,15.98=14,7\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{14,7.100}{10}=147\left(g\right)\)
Chúc bạn học tốt
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0 ,05 0,15
a)\(V=0,15\cdot22,4=3,36\left(l\right)\)
b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,7}{9,8}\cdot100=150\left(g\right)\)
c) \(m_{H_2}=0,15\cdot2=0,3\left(g\right)\)
\(m_{ddsau}=2,7+150-0,3=152,4\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)
\(\Rightarrow C\%=\dfrac{17,1}{152,4}\cdot100=11,22\%\)
\(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)
PTHH: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
________0,15------->0,3_________________________(mol)
=> \(m_{NaOH}=0,3.40=12\left(g\right)\)
=> \(m_{ddNaOH}=\dfrac{12.100}{10}=120\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{200.9,8\%}{98}=0,2\left(mol\right)\)
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,4 0,2
\(m_{ddNaOH}=\dfrac{0,4.40.100}{10}=160\left(g\right)\)
⇒ Chọn C