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Bài 5:
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
a, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
_____0,2__0,25__0,1 (mol)
b, VO2 = 0,25.22,4 = 5,6 (l)
c, PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
______0,1______________0,2 (mol)
\(\Rightarrow m_{H_3PO_4}=0,2.98=19,6\left(g\right)\)
\(\Rightarrow C\%_{H_3PO_4}=\dfrac{19,6}{120}.100\%\approx16,33\text{ }\%\)
Bạn tham khảo nhé!
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:
\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)
\(n_{P_2O_5}=\dfrac{99,4}{142}=0,7\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
0,7 2,1 1,4
a, \(m_{H_3PO_4}=1,4.98=137,2\left(g\right)\)
\(m_{ddH_3PO_4}=99,4+500=599,4\left(g\right)\)
Kl nước trong dd A :
\(m_{H_2O}=599,4-137,2=462,2\left(g\right)\)
\(b,C\%_{H_3PO_4}=\dfrac{137,2}{599,4}.100\%\approx22,89\%\)
\(c,C_M=\dfrac{n}{V}=\dfrac{1,4}{0,5}=2,8M\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,02\left(mol\right)\\n_{ZnCl_2}=0,01\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,02}{0,05}=0,4\left(M\right)\\m_{ZnCl_2}=0,01\cdot136=1,36\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
a.\(n_{NaOH}=\dfrac{8}{40}=0,2mol\)
\(V_{dd}=\dfrac{120}{1,2}=100ml=0,1l\)
\(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)
b.\(n_{NaOH}=\dfrac{21,6}{40}=0,54mol\)
\(V_{dd}=\dfrac{180}{1,2}=150ml=0,15l\)
\(C_{M_{NaOH}}=\dfrac{0,54}{0,15}=3,6M\)
\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{FeCl_2} = n_{Fe} = 0,1(mol)\\ m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ d) n_{HCl} = 2n_{Fe} = 0,2(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M\\ e)n_{Fe_3O_4} = \dfrac{2,32}{232} = 0,01(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ 4n_{Fe_3O_4} = 0,04 < n_{H_2} = 0,1 \to H_2\ dư\\ \)
\(n_{Fe} = 3n_{Fe_3O_4} = 0,03(mol)\\ m_{Fe} = 0,03.56 = 1,68(gam)\)
a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(m_{CuCl_2}=270.10\%=27\left(g\right)\Rightarrow n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)
Ta có: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\) ⇒ Fe hết, CuCl2 dư
PTHH: Fe + CuCl2 ---> FeCl2 + Cu
Mol: 0,15 0,15 0,15 0,15
\(a=m_{Cu}=0,15.64=9,6\left(g\right)\)
b, \(m_{dd.sau.pứ}=8,4+270-9,6=268,8\left(g\right)\)
\(m_{CuCl_2dư}=\left(0,2-0,15\right).135=6,75\left(g\right)\)
\(\left\{{}\begin{matrix}C\%_{CuCl_2dư}=\dfrac{6,75.100\%}{268,8}=2,51\%\\C\%_{FeCl_2}=\dfrac{0,15.127.100\%}{268,8}=7,09\%\end{matrix}\right.\)
c, \(V_{ddCuCl_2}=\dfrac{270}{1,35}=200\left(ml\right)=0,2\left(l\right)\)
\(\left\{{}\begin{matrix}C_{M_{CuCl_2dư}}=\dfrac{0,2-0,15}{0,2}=0,25M\\C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75M\end{matrix}\right.\)
bài 18 thì sao ạ?