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\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,1-------------->0,1---->0,05
=> VH2 = 0,05.22,4 = 1,12 (l)
mdd sau pư = 2,3 + 197,8 - 0,05.2 = 200 (g)
=> \(C\%=\dfrac{0,1.40}{200}.100\%=2\%\)
\(V_{dd}=\dfrac{200}{1,08}=\dfrac{5000}{27}\left(ml\right)=\dfrac{5}{27}\left(l\right)\)
=> \(C_M=\dfrac{0,1}{\dfrac{5}{27}}=0,54M\)
2NaOH + CuSO4 → Cu(OH)2 + Na2SO4
n NaOH = 0,2.5 = 1(mol)
n CuSO4 = 0,1.2 = 0,2(mol)
Ta có :
n NaOH / 2 = 0,5 > n CuSO4 / 1 = 0,2 => NaOH dư
n Cu(OH)2 = n CuSO4 = 0,2 mol
=> m A = 0,2.98 = 19,6 gam
n Na2SO4 = n CuSO4 = 0,2 mol
n NaOH pư = 2n CuSO4 = 0,4(mol)
V dd = 0,2 + 0,1 = 0,3(lít)
Suy ra:
CM Na2SO4 = 0,2/0,3 = 0,67M
CM NaOH = (1 - 0,4)/0,3 = 2M
Bài 9:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{14,6\%.100}{36,5}=0,4\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,4-2.0,1=0,2\left(mol\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)
a. Ta có: mNaOH=\(\frac{200.20}{100}=40\left(g\right)\)
pt : NaOH + HCl --------> NaCl + H2O
theo pt: 40g 36,5g 58,5g 18g
theo đề: 40g 36,5g 58,5g
=>\(C_{\%}=\frac{58,5}{200+100}.100\%=19,5\%\)
b.\(C_{\%}=\frac{36,5}{100}.100\%=36,5\%\)
\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)
\(n_{CuSO_4}=0.1\cdot2=0.2\left(mol\right)\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)
\(0.1.............0.05...............0.05...........0.05\)
\(m_{Cu\left(OH\right)_2}=0.05\cdot98=4.9\left(g\right)\)
\(C_{M_{Na_2SO_4}}=\dfrac{0.05}{0.2+0.1}=0.167\left(M\right)\)
\(C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0.2-0.05}{0.1}=1.5\left(M\right)\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{HCl}=0,15.4=0,6\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Zn}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)
c, \(n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\Rightarrow C_{M_{ZnCl_2}}=\dfrac{0,3}{0,15}=2\left(M\right)\)
Ta có : nNa \(=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH : \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\)
a) Theo pt :\(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b) Theo pt : \(n_{NaOH}=n_{Na}=0,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)
\(m_{ddspu}=4,6+100-0,1.2=104,4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{8}{104,4}.100\%=7,66\%\)