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\(a,PTHH:K_2O+H_2O\to 2KOH\\ n_{K_2O}=\dfrac{18,8}{94}=0,2(mol)\\ \Rightarrow n_{KOH}=0,4(mol)\\ \Rightarrow C\%_{KOH}=\dfrac{0,4.56}{18,8+121,2}.100\%=16\%\\ b,n_{KOH}=\dfrac{50.16\%}{56}=\dfrac{1}{7}(mol)\\ PTHH:2KOH+H_2SO_4\to K_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{2}{7}(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{\dfrac{2}{7}.98}{20\%}=140(g)\)
\(n_{K_2SO_4}=n_{KOH}=\dfrac{1}{7}(mol)\\ \Rightarrow m_{K_2SO_4}=\dfrac{1}{7}.174=24,86(g)\\ \Rightarrow C\%_{K_2SO_4}=\dfrac{24,86}{50+140}.100\%=13,08\%\)
Tham khảo
https://hoc247.net/cau-hoi-hoa-tan-naoh-ran-vao-nuoc-de-tao-thanh-2-dung-dich-a-va-b--qid95961.html
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2-->0,4----->0,2------->0,2
a
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b
\(CM_{MgCl_2}=\dfrac{0,2}{0,2}=1M\)
c
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
0,2------>0,4
\(V_{dd.NaOH}=\dfrac{0,4}{2}=0,2\left(l\right)\)
a)
\(SO_3 + H_2O \to H_2SO_4\)
Theo PTHH : \(n_{H_2SO_4} = n_{SO_2} = \dfrac{8}{80} = 0,1(mol)\)
\(\Rightarrow C\%_{H_2SO_4} = \dfrac{0,1.98}{200}.100\% = 4,9\%\)
b)
\(2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O\)
Theo PTHH : \(n_{NaOH} = 2n_{H_2SO_4} = 0,2(mol)\\ \Rightarrow m_{dd\ NaOH} = \dfrac{0,2.40}{4\%} = 200(gam)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
a+b) Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)=n_{H_2SO_4}=n_{ZnSO_4}\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5\left(M\right)=C_{M_{ZnSO_4}}\)
c) Theo PTHH: \(n_{H_2}=n_{Zn}=0,3mol\) \(\Rightarrow V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\)
d) Theo PTHH: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2mol\)
\(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\)
a) nH2SO4 = 0,2 . 1 = 0,2 mol
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
0,2 0,4
mNaOH = 0,4 . 40 = 16g
mddNaOH = \(\frac{16.100\%}{20\%}=80g\)
b) 2KOH + H2SO4 -> K2SO4 + 2H2O
0,4 <---------- 0,2
=> mKOH = 0,4 . 56 = 22,4 g
mddKOH = \(\frac{22,4.100\%}{5,6\%}=400g\)
VddKOH = \(\frac{400}{1,045}=383ml\)
\(A+H_2O\rightarrow AOH+\dfrac{1}{2}H_2\)
Bảo toàn khối lượng => \(m_{H_2}=3,45+102,7-106=0,15\left(g\right)\)
=> \(n_A=2n_{H_2}=0,15\left(mol\right)\)
=>\(M_A=\dfrac{3,45}{0,15}=23\)
=> A là Na
\(C\%_{NaOH}=\dfrac{0,15.40}{106}.100=5,66\%\)