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\(a.V_{CO_2\left(dktc\right)}=0,25.22,4=5,6\left(l\right)\)
\(b.m_{Al_2O_3}=0,5.160=80\left(g\right)\)
4.
a) \(V_{SO_2}=0.5\cdot22.4=11.2\left(l\right)\)
b) \(V_{CH_4}=\dfrac{3.2}{16}\cdot22.4=4.48\left(l\right)\)
c) \(V_{N_2}=\dfrac{0.9\cdot10^{23}}{6\cdot10^{23}}\cdot22.4=3.36\left(l\right)\)
5.
a) \(m_{Al}=0.1\cdot27=2.7\left(g\right)\)
b) \(m_{Cu\left(NO_3\right)_2}=0.3\cdot188=56.4\left(g\right)\)
c) \(m_{Na_2CO_3}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}\cdot106=21.2\left(g\right)\)
d) \(m_{CO_2}=\dfrac{8.96}{22.4}\cdot44=17.6\left(g\right)\)
e) \(m_K=0.5\cdot2\cdot39=39\left(g\right)\\ m_C=0.5\cdot12=6\left(g\right)\\ m_O=0.5\cdot3\cdot16=24\left(g\right)\)
\(a.m_O=1.16=16\left(g\right)\\ m_{O_2}=1.32=32\left(g\right)\\ b.m_{Fe}=1,5.56=84\left(g\right)\\ m_{Fe_2O_3}=1,5.160=240\left(g\right)\\ c.m_N=0,25.14=3,5\left(g\right)\\ m_{NO_2}=2,5.46=115\left(g\right)\\ d.m_{C_6H_{12}O_6}=1.180=180\left(g\right)\)
a) nCO2=[(9.1023)/(6.1023)]=1,5(mol)
=> mCO2=1,5.44=66(g)
V(CO2,đktc)=1,5.22,4=33,6(l)
b) nH2=4/2=2(mol)
N(H2)=2.6.1023=12.1023(phân tử)
V(H2,đktc)=2.22,4=44,8(l)
c) N(CO2)=0,5.6.1023=3.1023(phân tử)
V(CO2,đktc)=0,5.22,4=11,2(l)
mCO2=0,5.44=22(g)
d) nN2=2,24/22,4=0,1(mol)
mN2=0,1.28=2,8(g)
N(N2)=0,1.1023.6=6.1022 (phân tử)
e) nCu=[(3,01.1023)/(6,02.1023)]=0,5(mol)
mCu=0,5.64=32(g)
Mà sao tính thể tích ta :3
Câu 1:
\(m_{H_2S}=0,75.34=25,5(g)\\ m_{CaSO_4}=0,025.136=3,4(g)\\ m_{Fe_2O_3}=0,05.160=8(g)\)
Câu 2:
\(V_{N_2}=2,5.22,4=56(l)\\ V_{H_2}=0,03.22,4=0,672(l)\\ V_{O_2}=0,45.22,4=10,08(l)\\ V_{hh}=22,4.(0,2+0,25)=22,4.0,45=10,08(l)\)
\(n_{H_2O}=\dfrac{24.10^{23}}{6.10^{23}}=4\left(mol\right)\\ n_{CO_2}=\dfrac{1,44.10^{23}}{6.10^{23}}=0,24\left(mol\right)\\ n_{Fe}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\\ n_C=\dfrac{0,66.10^{23}}{6.10^{23}}=0,11\left(mol\right)\)
a) \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)=>V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)=>V_{O_2}=0,1.22,4=2,24\left(l\right)\)
b) \(n_{H_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
=> mH2O = 0,5.18 = 9(g)
c) \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
=> Số nguyên tử Mg = 0,5.6.1023 = 3.1023
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
=> Số nguyên tử Zn = 0,2.6.1023 = 1,2.1023
Số nguyên tử Ag = 0,15.6.1023 = 0,9.1023
Số nguyên tử Al = 0,45.6.1023 = 2,7.1023
