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Câu 14 :
\(n_{H_2SO_4}=0.15\cdot2=0.3\left(mol\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(0.6..........0.3\)
\(m_{KOH}=0.3\cdot56=33.6\left(g\right)\)
\(m_{dd_{KOH}}=\dfrac{33.6}{40\%}=84\left(g\right)\)
Câu 13 :
\(n_{HCl}=2\cdot0.6=1.2\left(mol\right)\)
\(\Rightarrow n_{H_2O}=\dfrac{1}{2}\cdot1.2=0.6\left(mol\right)\)
Bảo toàn nguyên tố O :
\(n_{O\left(oxit\right)}=n_{H_2O}=0.6\left(mol\right)\)
\(\Rightarrow m_O=0.6\cdot16=9.6\left(g\right)\)
\(m_{Fe}=34.8-9.6=25.2\left(g\right)\)
\(n_{Fe}=\dfrac{25.2}{56}=0.45\left(mol\right)\)
\(n_{Fe}:n_O=0.45:0.6=3:4\)
\(CT:Fe_3O_4\)
\(m_{Muối}=m_{oxit}+m_{HCl}-m_{H_2O}=34.8+1.2\cdot36.5-0.6\cdot18=67.8\left(g\right)\)
\(a.n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.0,1........0,2..........0,1.........0,1\left(mol\right)\\ C\%_{ddH_2SO_4}=\dfrac{0,1.98}{140}.100=7\%\\ c.m_{dd.muối}=6,5+140-0,1.2=146,3\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,1}{146,3}.100\approx9,296\%\\ \)
a.nZn=6,565=0,1(mol)Zn+2HCl→ZnCl2+H2b.0,1........0,2..........0,1.........0,1(mol)C%ddH2SO4=0,1.98140.100=7%c.mdd.muối=6,5+140−0,1.2=146,3(g)C%ddZnCl2=136.0,1146,3.100≈9,296%
\(\%Fe_{FeCl_3}=\dfrac{56}{56+35.5\cdot3}\simeq34,46\%\)
Fe không có trong CuSO4 nha bạn
\(M_{KNO_3}=39+14+16\cdot3=101g/mol\\ M_{H_2SO_4}=1\cdot2+32+16\cdot4=98g/mol\\ M_{CuSO_4}=64+32+16\cdot4=160g/mol\\ M_{Al\left(OH\right)_3}=27+\left(1+17\right)\cdot3=78g/mol\\ M_{NaCl}=23+35,5=58,5g/mol\)
Bài 8 :
200ml = 0,2l
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
Pt : \(R+2HCl\rightarrow RCl_2+H_2|\)
1 2 1 1
0,1 0,2 0,1
a) \(n_{H2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) \(n_R=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(M_R=\dfrac{2,4}{0,1}=24\) (g/mol)
Vậy kim loại R là magie
Chúc bạn học tốt
a)\(2Ca+O_2\underrightarrow{t^o}2CaO\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)
\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)