Ta có:

+ \(M_{CO_2}=12+16.2=44\) g/mol

⇒ \(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=\dfrac{1}{4}mol\)

+ \(n=\dfrac{sophantu}{6.10^{23}}=\dfrac{9.10^{23}}{6.10^{23}}=\dfrac{3}{2}=mol\)

\(V_{H_2\left(đktc\right)}=n.22,4=\dfrac{3}{2}.22.4=33,6\left(l\right)\)

\(a.m_{CuSO_4}=n.M=0,3.160=48\left(g\right)\)

\(b.n_{CaCO_3}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ m_{CaCO_3}=n.M=1,5.100=150\left(g\right)\)

\(c.n_{MgCl_2}=\dfrac{1,5.10^{22}}{6.10^{23}}=0,025\left(mol\right)\\ \Rightarrow m_{MgCl_2}=n.M=0,025.95=2,375\left(g\right)\)

V C O 2 = n C O 2  . 22,4 = 0,25.22,4 = 5,6 (l)

V O 2 = n O 2 .22,4 = 0,25.22,4 = 5,6 (l)

n N 2 == 0,75(mol)

→ V N 2 = n N 2 .22,4 = 0,75.22,4= 16,8 (l)

n C O 2 == 0,2 (mol)

→ V C O 2 = n C O 2 . 22,4 = 0,2.22,4 = 4,48 (l)

n H 2 ==1,5(mol)

→ V H 2 = n H 2 . 22,4 = 1,5. 22,4 = 33,6 (l)

n C O == 0,05(mol)

→ V C O = n C O . 22,4 = 0,05. 22,4 = 1,12 (l)

\(m_{CO_2}=12+16\cdot2=44\)

Số mol của CO₂

\(n=\frac{m}{M}=\frac{11}{44}=0,25\)(mol)

Số mol của H₂

\(n=\frac{sophantu}{6.10^{23}}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(\Rightarrow\) V_H2(đktc) =n.22,4=1,5.22,4=33,6(lít)

=> V_H2(đktc) = 1,5 x 22,4 = 33,6 lít

 mCO2 = 12 + 16.2 = 44
Số mol của CO2
n = m/M = 11/44 = 0,25(mol)
Số mol của H2
n = số phần tử/6.10^23 = 9.10^23/6.10^23 = 1,5(mol)
=> VH2(đktc) = n.22,4 = 1,5.22,4 = 33,6(lít)

+ \(M_{CO_2}=12+16.2=44\left(\dfrac{g}{mol}\right)\)

\(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25mol\)

+ \(n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5mol\)

\(V_{H_2\left(đktc\right)}=n.22,4=1,5.22,4=33,6\left(l\right)\)

Câu 2

\(n_{O_2}=\dfrac{V_{O_2\left(đktc\right)}}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Số phân tử oxi \(=\)\(0,3.6.10^{23}=1,8.10^{23}\)

\(M_{O_2}=16.2=32\left(\dfrac{g}{mol}\right)\)

\(m_{O_2}=n.M=0,3.32=9,6g\)

a. \(m_{CO_2}=\dfrac{8,96}{22,4}.44=17,6\left(g\right)\)

b. \(m_{Fe}=\dfrac{1,8.10^{22}}{6.10^{23}}.56=1,68\left(g\right)\)

c. \(m_{Fe_2\left(SO_4\right)_3}=0,25.400=100\left(g\right)\)

\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)

V_CO2=0,5.22,4=1,12l

V_Cl2=0,25.22,4=5,6l

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