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a) Ta có: \(n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{11}{44}=0,25\left(mol\right)\)
\(V_{CO_2}=n_{CO_2}.22,4=0,25.22,4=5,6\left(l\right)\)
b) Ta có: \(n_{Fe_2O_3}=\frac{m_{Fe_2O_3}}{M_{Fe_2O_3}}=\frac{80}{160}=0,5\left(mol\right)\)
a) nCO2 = mCO2 : MCO2 = 11 : 18 = 0,6 (mol)
=> VCO2 = nCO2 * 22,4 = 0,6 * 22,4 = 13,44 (lít)
b) nFe2O3 = mFe2O3 : MFe2O3 = 80 : 160 = 0,5 (mol)
a) mFeSO4= 0,25.152=38(g)
b) mFeSO4= \(\dfrac{13,2.10^{23}}{6.10^{23}}.152=334,4\left(g\right)\)
c) mNO2= \(\dfrac{8,96}{22,4}.46=18,4\left(g\right)\)
d) mA= 27.0,22+64.0,25=21,94(g)
e) mB= \(\dfrac{11,2}{22,4}.32+\dfrac{13,44}{22,4}.28=32,8\left(g\right)\)
g) mC= \(64.0,25+\dfrac{15.10^{23}}{6.10^{23}}.56=156\left(g\right)\)
h) mD= \(0,25.32+\dfrac{11,2}{22,4}.44+\dfrac{2,7.10^{23}}{6.10^{23}}.28=42,6\left(g\right)\)
hơi muộn nha<3
a,Đặt tạm Al2(SO4)3 là A nhé 
Có: nA=\(\frac{m_A}{M_A}\)=\(\frac{75,24}{27.2+\left(32+16.4\right).3}\)=0,22(mol)
b,Tương tự: nA=\(\frac{V_{A\left(Đktc\right)}}{22,4}\)=0.7(mol)
nSO2 = 20,26 / 22,4 = 0,9 (mol)
nCO = \(\frac{7,2.10^{23}}{6.10^{23}}=1,2\left(mol\right)\)
=> Khối lượng hỗn hợp là:
mhỗn hợp = 5,6 + 0,9 x 64 + 1,2 x 28 + 0,5 x 32 = 112,8 (gam)
Bài 5:
Ta có mS = 0,5 . 32 = 16 ( gam )
=> mFe = 56 . 0,6 = 33,6 ( gam )
=> mFe2O3 = 160 . 0,8 = 128 ( gam )
=> mhỗn hợp = 16 + 33,6 + 128 = 177,6 ( gam )
1.
Gọi CTHH của HC là A2O3
A2O3 + 3H2SO4 →→A2(SO4)3 + 3H2O
mH2SO4=12,25.24100=2,94(g)12,25.24100=2,94(g)
nH2SO4=2,9498=0,03(mol)2,9498=0,03(mol)
Theo PTHH ta có:
1313nH2SO4=nA2O3=0,01(mol)
MA2O3=1,020,01=1021,020,01=102
MA=102−16.32=27102−16.32=27
Vậy A là Al,CTHH của oxit là Al2O3
2.
a;
-nO2=2,4.10236.1023=0,4(mol)2,4.10236.1023=0,4(mol)⇒⇒VO2=22,4.0,4=8,96(lít)
-nCO2=9.10236.1023=1,5(mol)9.10236.1023=1,5(mol)⇒⇒VCO2=1,5.22,4=33,6(lít)
-nO2=6,432=0,2(mol)6,432=0,2(mol);nN2=22,428=0,8(mol)22,428=0,8(mol)⇒⇒Vhh=(0,8+0,2).22,4=22,4(lí t)(0,8+0,2).22,4=22,4(lít)
-Vhh=(0,75+0,5+0,25).22,4=33,6(lí t)(0,75+0,5+0,25).22,4=33,6(lít)
b;
-mNaOH=40.2=80(g)
-nMg=1,32.10226.1023=0,022(mol)1,32.10226.1023=0,022(mol)⇒⇒mMg=0,022.24=0,528(g)
-nHCl=9,6.10226.1023=0,16(mol)⇒9,6.10226.1023=0,16(mol)⇒mHCl=36,5.0,16=5,84(g)
-nCO2=3322,4=165112(mol)3322,4=165112(mol)
nCO=11,222,4=0,5(mol)11,222,4=0,5(mol)
nN2=5,522,4=55224(mol)5,522,4=55224(mol)
mhh=44.165112165112+28.0,5+28.5522455224=85,7(g)
1)\(\text{mH2SO4=0,2.98=19,6(g)}\)
nHCl=\(\frac{0,3.10^{23}}{6.10^{23}}\)=0,05(mol)
\(\text{mHCl=36,5.0,05=1,825(g)}\)
nN2=\(\frac{6,72}{22,4}\)=0,3(mol)
\(\text{mN2=0,3.28=8,4(g)}\)
2)
nCO2=\(\frac{0,44}{44}\)=0,01(mol)
nH2=\(\frac{0,04}{2}\)=0,02(mol)
nN2=\(\frac{0,56}{28}\)=0,02(mol)
\(\text{nhh=0,01+0,02+0,02=0,05(mol)}\)
\(\text{Vhh=0,05.22,4=1,12(l)}\)
Số phân tử=0,05.6.1023=5.1021( phân tử)
a)
nSO2=\(\frac{2,24}{22,4}\)=0,1(mol)
mSO2=0,1.64=6,4(g)
Số phân tử SO2=0,1.6.1023=6.1022( phân tử)
b)
nSO3=\(\frac{4}{80}\)=0,05(mol)
số phân tử SO3=0,05.6.1023=3.1022(phân tử)
VSO2=0,05.2,4=1,12(l)
c)
nCl2=3.1023/6.1023=0,5(mol)
VCl2=0,5.22,4=11,2(l)
mCl2=0,5.71=35,5 g
d)
nC2H4=\(\frac{3,36}{22,4}\)=0,15(mol)
Số phân tử=0,15.6.1023=9.1022
mC2H4=0,15.28=4,2 g
\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)\)
\(n_{CH_4}=\dfrac{6\times10^{22}}{6\times10^{23}}=0,1\left(mol\right)\)
\(\Rightarrow V_{hh}=\left(0,3+0,25+0,1+0,2\right)\times22,4=19,04\left(l\right)\)
\(m_{Cl_2}=0,2\times71=11,2\left(g\right)\)
\(m_{SO_2}=0,3\times64=19,2\left(g\right)\)
\(m_{CH_4}=0,1\times16=1,6\left(g\right)\)
\(\Rightarrow m_{hh}=11,2+8+19,2+1,6=40\left(g\right)\)