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Bài 14:
a) \(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
PTHH: Ca + 2H2O --> Ca(OH)2 + H2
0,5<--------------0,5<----0,5
=> mCa = 0,5.40 = 20 (g)
=> \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{20}{34}.100\%=58,82\%\\\%m_{CaO}=100\%-58,82\%=41,18\%\end{matrix}\right.\)
b) b phải là khối lượng bazo thu được chứ nhỉ..., sao tính đc m dung dịch
\(n_{CaO}=\dfrac{34-20}{56}=0,25\left(mol\right)\)
PTHH: CaO + H2O --> Ca(OH)2
0,25---------->0,25
=> mCa(OH)2 = (0,5 + 0,25).74 = 55,5 (g)
\(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
0,5 0,5 0,5 ( mol )
( \(CaO+H_2O\) không giải phóng \(H_2\) )
\(m_{Ca}=0,5.40=20g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{20}{34}.100=58,82\%\\\%m_{CaO}=100\%-58,82\%=41,18\%\end{matrix}\right.\)
\(n_{CaO}=\dfrac{34-20}{56}=0,25\left(mol\right)\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
0,25 0,25 ( mol )
\(m_{Ca\left(OH\right)_2}=\left(0,5+0,25\right).74=55,5g\)
Bài 15:
a) \(n_{O_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
PTHH: 2H2O --đp--> 2H2 + O2
1<--------------0,5
=> \(H=\dfrac{1.18}{22,5}.100\%=80\%\)
b) \(n_{H_2O}=\dfrac{81}{18}=4,5\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
4,5<------------4,5
=> \(V_{H_2\left(lý.thuyết\right)}=4,5.24,79=111,555\left(l\right)\)
=> \(V_{H_2\left(tt\right)}=\dfrac{111,555.100}{90}=123,95\left(l\right)\)
c) \(n_{H_2}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)
PTHH: 2H2O --đp--> 2H2 + O2
1,25<-------1,25
=> \(m_{H_2O\left(lý.thuyết\right)}=1,25.18=22,5\left(g\right)\)
=> \(m_{H_2O\left(tt\right)}=\dfrac{22,5.100}{75}=30\left(g\right)\)
\(n_{H_2O}=\dfrac{22,5}{18}=1,25\left(mol\right)\)
\(n_{O_2}=\dfrac{12,395}{24,79}=0,5mol\)
\(2H_2O\rightarrow\left(điện.phân\right)2H_2+O_2\)
1,25 0,5 ( mol ) ( thực tế )
1 0,5 ( mol ) ( lý thuyết )
\(H=\dfrac{1}{1,25}.100=80\%\)
b.\(n_{H_2O}=\dfrac{81}{18}=4,5\left(mol\right)\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
4,5 4,5 ( mol )
\(V_{H_2}=4,5.24,79:90\%=123,95l\)
c.\(n_{H_2}=\dfrac{30,9875}{24,79}=1,25mol\)
\(2H_2O\rightarrow\left(điện.phân\right)2H_2+O_2\)
1,25 1,25 ( mol )
\(m_{H_2O}=1,25.18:75\%=30g\)
Câu 1 :
$n_C = \dfrac{4,8}{12} = 0,4(mol) ; n_{O_2} = \dfrac{7,437}{24,79} = 0,3(mol)$$
$C + O_2 \xrightarrow{t^o} CO_2$
Ta thấy :
$n_C : 1 > n_{O_2} : 1$ nên C dư
$n_{C\ pư} = n_{O_2} = 0,3(mol) \Rightarrow m_{C\ dư} = (0,4 - 0,3).12 = 1,2(gam)$
$\Rightarorw V_{CO_2} = V_{O_2} = 7,437(lít)$
Câu 2 :
$n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)$
$n_{Cl_2} = \dfrac{9,916}{24,79} = 0,4(mol)$
$Mg + Cl_2 \xrightarrow{t^o} MgCl_2$
Ta thấy :
$n_{Mg} : 1 < n_{Cl_2} : 1$ nên $Cl_2$ dư
$n_{Cl_2\ pư} = n_{Mg} = 0,1(mol) \Rightarrow m_{Cl_2\ dư} = (0,4 - 0,1).71 = 21,3(gam)$
$n_{MgCl_2}= n_{Mg} = 0,1(mol) \Rightarrow m_{MgCl_2} = 0,1.95 = 9,5(gam)$
1. a) \(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)
b)PT: 214----------------->160 (kg)
Đề: x------------------------>80(kg)
=> x=\(\dfrac{80.214}{160}=107\left(kg\right)\)
c) => \(H=\dfrac{107}{120}.100=89,17\%\)
2. a) \(CaCO_3-^{t^o}\rightarrow CaO+CO_2\)
\(n_{CaCO_3\left(pứ\right)}=n_{CaO}=\dfrac{151,2}{56}=2,7\)
=> \(m_{CaCO_3\left(pứ\right)}=2,7.100=270\left(kg\right)\)
=> \(H=\dfrac{270}{300}.100=90\%\)
b) \(n_{CO_2}=n_{CaO}=\dfrac{151200}{56}=2700\left(mol\right)\)
=> \(V_{CO_2}=2700.22,4=60480\left(l\right)\)
a) CH4 + 2O2 --to--> CO2 + 2H2O
b) \(V_{O_2}=\dfrac{56}{5}=11,2\left(l\right)\)
=> \(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
_____0,25<--0,5-------->0,25
=> VCH4 = 0,25.22,4 = 5,6 (l)
c)
mCO2 = 0,25.44 = 11 (g)
n C3H8=0,1 mol
C3H8+5O2-to>3CO2+4H2O
0,1-----0,5------------0,3-----0,4 mol
=>VCO2=0,3.22,4=6,72 l
=>VO2=0,5.22,4=11,2l
Sửa đề: Cho \(65g\) kẽm
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=\dfrac{24,79}{24,79}=1(mol)\\ \Rightarrow m_{H_2}=1.2=2(g)\\ \text {Bảo toàn KL}:m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=136+2-65=73(g)\)
\(n_{ZnCO_3}=\dfrac{75}{125}=0,6\left(mol\right);n_{CO_2\left(TT\right)}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\\ ZnCO_3\rightarrow\left(t^o\right)ZnO+CO_2\\ n_{CO_2\left(LT\right)}=n_{ZnCO_3}=0,6\left(mol\right)\\ H=\dfrac{0,5}{0,6}.100\%\approx83,333\%\)
em cảm ơn ạ