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Bài 9:
a) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,2->0,25------> 0,1
=> mP2O5 = 0,1.142 = 14,2 (g)
b)
P2O5 + 3H2O --> 2H3PO4
dd H3PO4 là dd axit nên quỳ tím đổi màu đỏ
a) \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,1--------------->0,05
=> mP2O5 = 0,05.142 = 7,1 (g)
b)
PTHH: P2O5 + 3H2O --> 2H3PO4
dd B là dd axit nên quỳ tím chuyển màu đỏ
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2 0,2 0,2 0,2
a)\(V_{H_2}=0,2\cdot22,4=4,48l\)
b)\(m_{ZnSO_4}=0,2\cdot161=32,2g\)
\(m_{ddZnSO_4}=30+200-0,2\cdot2=229,6g\)
\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{32,2}{229,6}\cdot100\%=14,02\%\)
c)\(n_{CuO}=\dfrac{24}{80}=0,3mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,3 0,2 0,2
\(m_{rắn}=m_{Cu}=0,2\cdot64=12,8g\)
nZn=1365=0,2molnZn=1365=0,2mol
Zn+H2SO4→ZnSO4+H2Zn+H2SO4→ZnSO4+H2
0,2 0,2 0,2 0,2
a)VH2=0,2⋅22,4=4,48lVH2=0,2⋅22,4=4,48l
b)mH2SO4=0,2⋅98=19,6gmH2SO4=0,2⋅98=19,6g
C%=mctmdd⋅100%=19,6200⋅100%=9,8%C%=mctmdd⋅100%=19,6200⋅100%=9,8%
c)nCuO=2480=0,3molnCuO=2480=0,3mol
CuO+H2→Cu+H2OCuO+H2→Cu+H2O
0,3 0,2 0,2
mrắn=mCu=0,2⋅64=12,8g.
\(n_K=\dfrac{39}{39}=1\left(mol\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{KOH}=n_K=1\left(mol\right)\\ C_{MddKOH}=\dfrac{1}{0,2}=5\left(M\right)\\ c,2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)
nNa = 6.9 : 23 = 0.3 mol
4Na + O2 ->2 Na2O
mol : 0.3 -> 0.15
Na2O + H2O -> 2NaOH
mol : 0.15 -> 0.3
mdd = 0.15 x 62 + 140.7 = 150g
C% NaOH = 0.3x40: 150 x 100% = 8%
\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\\ n_{H_2O}=\dfrac{15}{18}=\dfrac{5}{6}\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
LTL: \(2>\dfrac{5}{6}\) => Na dư
Theo pthh: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1}{2}n_{H_2O}=\dfrac{1}{2}.\dfrac{5}{6}=\dfrac{5}{12}\left(mol\right)\\n_{Na\left(pư\right)}=n_{NaOH}=n_{H_2O}=\dfrac{5}{6}\left(mol\right)\end{matrix}\right.\)
=> \(V_{H_2}=\dfrac{5}{12}.22,4=\dfrac{28}{3}\left(l\right)\)
\(m_{dd}=15+23.\dfrac{5}{6}-\dfrac{5}{12}.2=\dfrac{100}{3}\\ m_{NaOH}=\dfrac{5}{6}.40=\dfrac{100}{3}\left(g\right)\\ \rightarrow C\%_{NaOH}=\dfrac{\dfrac{100}{3}}{\dfrac{100}{3}}.100\%=100\%\)
\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\\ n_{H_2O}=\dfrac{15}{18}=0,83\left(mol\right)\\ pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,83 0,83 0,416
\(V_{H_2}=0,416.22,4=9,3l\\
m_{\text{dd}}=46+15-\left(0,416.2\right)=60,17\left(g\right)C\%=\dfrac{0,83.40}{60,17}.100\%=55,176
\%\)
\(2RS+3O_2\underrightarrow{^{^{t^0}}}2RO+2SO_2\)
\(RO+H_2SO_4\rightarrow RSO_4+H_2O\)
Giả sử :
\(n_{H_2SO_4}=1\left(mol\right)\)
\(\Rightarrow m_{dd_{H_2SO_4}}=\dfrac{98}{24.5\%}=400\left(g\right)\)
\(m_{\text{dung dịch muối}}=R+16+400=R+416\left(g\right)\)
\(C\%_{RSO_4}=\dfrac{R+96}{R+416}\cdot100\%=33.33\%\)
\(\Rightarrow R=64\)
\(R:Cu\)
\(n_{CuS}=\dfrac{12}{96}=0.125\left(mol\right)\)
\(n_{CuSO_4}=n_{CuS}=0.125\left(mol\right)\)
\(m_{CuSO_4}=0.125\cdot160=20\left(g\right)\)
\(m_{dd}=0.125\cdot80+\dfrac{0.125\cdot98}{24.5\%}=60\left(g\right)\)
Khối lượng dung dịch bão hòa còn lại :
\(60-15.625=44.375\left(g\right)\)
\(CT:CuSO_4\cdot nH_2O\)
\(m_{CuSO_4}=m\left(g\right)\)
\(C\%=\dfrac{m}{44.375}\cdot100\%=22.54\%\)
\(\Rightarrow m=10\)
\(m_{CuSO_4\left(tt\right)}=20-10=10\left(g\right)\)
\(\dfrac{10}{15.625}=\dfrac{160}{M_{tt}}\)
\(\Rightarrow M_{tt}=250\)
\(\Rightarrow n=5\)
\(CT:CuSO_4\cdot5H_2O\)
a) Y là Cu
$m_{Cu} = 8(gam)$
Gọi $n_{Al} = a(mol) ; n_{Fe} = b(mol)$
Ta có : $27a + 56b + 8 = 13,45(1)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2} = 1,5a + b = \dfrac{5,6}{22,4} = 0,25(2)$
Từ (1)(2) suy ra a = 0,15 ; b = 0,025$
$\%m_{Cu} = \dfrac{8}{13,45}.100\% = 59,47\%$
$\%m_{Al} = \dfrac{0,15.27}{13,45}.100\% = 30,11\%$
$\%m_{Fe} = 10,42\%$
b)
$n_{H_2SO_4} = n_{H_2} = 0,25(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,25}{0,5} = 0,5(lít)$
