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nH2= 0,56 : 22,4 = 0,025 mol
mH2= 0,025 . 2=0,05g
nhcl=6,72:22,4=0,2 mol
mhcl=0,2 . 36,5 = 7,3g
a) \(n_{H_2}=\frac{0,56}{22,4}=0,025\left(mol\right)\)
\(\Rightarrow m_{H_2}=n.M=0,025\times2=0,05\left(gam\right)\)
b) \(n_{HCl}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl}=n.M=0,3\times36,5=10,95\left(gam\right)\)
a.
\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)
\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)
\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)
\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)
b.
\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)
a, khối lượng của 2,5 mol CuO là:
\(m=n.M=2,5.80=200\left(g\right)\)
b, số mol của 4,48 lít khí CO2 (đktc) là:
\(n=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Giả sử có 1 mol khí Cl2, 2 mol khí O2
a) \(\left\{{}\begin{matrix}\%V_{Cl_2}=\dfrac{1}{1+2}.100\%=33,33\%\\\%V_{O_2}=\dfrac{2}{1+2}.100\%=66,67\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Cl_2}=\dfrac{1.71}{1.71+2.32}.100\%=52,59\%\\\%m_{O_2}=\dfrac{2.32}{1.71+2.32}.100\%=47,41\%\end{matrix}\right.\)
b) \(\overline{M}=\dfrac{1.71+2.32}{1+2}=45\left(g/mol\right)\)
=> \(d_{A/H_2}=\dfrac{45}{2}=22,5\)
c) \(n_A=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
=> mA = 0,3.45 = 13,5 (g)
a, mCaO = 0,5.56 = 28 (g)
b, \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
c, \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)
d, \(V_{hhk}=0,2.22,4+0,3.22,4=11,2\left(l\right)\)
e, \(\%m_{Cu}=\dfrac{64}{64+32+16.4}.100\%=40\%\)
Bạn tham khảo nhé!
a) mCaO=nCaO.M(CaO)=0,5.56=28(g)
b) nCO2=V(CO2,dktc)=6,72/22.4=0,3(mol)
c) nH2SO4=mH2SO4/M(H2SO4)=24,5/98=0,25(mol)
d) V(hh H2,NH3)=(0,3+0,2).22,4=11,2(l)
e) %mCu/CuSO4=(64/160).100=40%
Chúc em học tốt!
\(a)\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3.2 = 0,6(mol)\\ \Rightarrow m_{KMnO_4} = 0,6.158 = 94,8(gam)\\ b)\ n_{KClO_3} = \dfrac{24,5}{122,5} = 0,2(mol)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{O_2} = \dfrac{3}{2}n_{KClO_3} = 0,3(mol)\\ \Rightarrow m_{O_2} = 0,3.32 = 9,6(gam)\)
a) nO2 = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Pt: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
Theo pt: \(n_{KMnO_4}=2nO_2=0,6\left(mol\right)\Rightarrow m_{KMnO_4}=0,6.158=94,8g\)
b) \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,3\left(mol\right)\)
\(Pt:2KClO_3\rightarrow2KCl+3O_2\)
\(nO_2=\dfrac{3}{2}n_{KClO_3}=0,45\left(mol\right)\Rightarrow mO_2=0,45.32=14,4g\)
N2+3H2to,xt,p⇌2NH3N2+3H2⇌to,xt,p2NH3
Ta thấy : VN2:1>VH2:3VN2:1>VH2:3 nên N2N2 dư
Gọi hiệu suất là a
Suy ra : VH2 pư=11,2a(lít)VH2 pư=11,2a(lít)
VN2 pư=11,2a3VN2 pư=11,2a3
VNH3=2.11,2a3VNH3=2.11,2a3
Ta có :
a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,2<----0,3<-----------------0,3
=> mAl = 0,2.27 = 5,4 (g)
c) \(m_{dd.H_2SO_4}=\dfrac{0,3.98}{30\%}=98\left(g\right)\)
mhh=\(\dfrac{3,36}{22,4}.64+\dfrac{2,8}{22,4}.28+\dfrac{6,72}{22,4}.2=13,7gam\)
=> ý A
D
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(Mol)\\ \Rightarrow m_{H_2}=0,3.2=0,6(g)\)