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\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=\dfrac{m_1}{23}+m_2-\dfrac{m_1}{46}=\dfrac{m_1}{46}+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{m_1}{46}+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=m_1+m_2-\dfrac{m_1}{23}=\dfrac{22}{23}m_1+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{22}{23}m_1+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
`C1:`
`2NaOH+H_2 SO_4 ->Na_2 SO_4 +2H_2 O`
`n_[H_2 SO_4]=0,2.1=0,2(mol)`
`n_[NaOH]=[200.10]/[100.40]=0,5(mol)`
Ta có: `[0,2]/1 < [0,5]/2=>NaOH` dư, `H_2 SO_4` hết.
`=>` Quỳ tím chuyển xanh.
`C2:`
`SO_3 +H_2 O->H_2 SO_4`
`0,2` `0,2` `(mol)`
`n_[SO_3]=16/80=0,2(mol)`
`C_[M_[H_2 SO_4]]=[0,2]/[0,25]=0,8(M)`
\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)
a)
C% CuSO4 = 16/(16 + 184) .100% = 8%
b)
n NaOH = 20/40 = 0,5(mol)
CM NaOH = 0,5/4 = 0,125M
a) \(n_{NaCl}=2,5.0,9=2,25\left(mol\right)\Rightarrow m_{NaCl}=2,25.58,5=131,625\left(g\right)\)
b) \(m_{MgCl_2}=\dfrac{50.4}{100}=2\left(g\right)\)
c) \(n_{MgSO_4}=0,25.0,1=0,025\left(mol\right)\Rightarrow m_{MgSO_4}=0,025.120=3\left(g\right)\)
d) \(m_{NaOH}=\dfrac{20.40}{100}=8\left(g\right)\)
Có lẽ bạn hiểu nhầm M (mol/l) với mol rồi :)
Sửa hết mol ---> M nha
\(a,n_{NaCl}=2,5.0,9=2,25\left(mol\right)\\ \rightarrow m_{NaCl}=2,25.58,5=131,625\left(g\right)\\ b,m_{MgCl_2}=\dfrac{4.50}{100}=2\left(g\right)\\ c,Đổi:250ml=0,25l\\ \rightarrow n_{MgSO_4}=0,1.0,25=0,025\left(mol\right)\\ \rightarrow m_{MgSO_4}=0,025.120=3\left(g\right)\\ d,m_{NaOH}=\dfrac{40.20}{100}=8\left(g\right)\)
Bài 1:
\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)
Bài 2:
\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)
1) m dd naOH (bđ) = \(250.1,2=300\left(g\right)\)
\(m_{NaOH\left(bđ\right)}=\frac{300.10}{200}=30\left(g\right)\)
\(n_{Na}=\frac{m}{23}\left(mol\right)\)
PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)
_______ \(\frac{m}{23}\) -------------------> \(\frac{m}{23}\)-------> \(\frac{m}{46}\)(mol)
\(=>m_{NaOH}=\frac{m}{23}.40=\frac{40m}{23}\left(g\right)\)
=> \(m_{dd}\) sau pư = \(m+300-2.\frac{m}{46}\) = \(\frac{22m}{23}+300\left(g\right)\)
=> \(C\%=\frac{\frac{40m}{23}}{\frac{22m}{23}+300}.100\%=14,94\%\)
=> m = 28,1 (g)