a) n Al2(so₄)3_{=\(\frac{34,2}{27.2+\left(32+16.4\right).3}\)}=0,1

b)n caco₃=\(\frac{12}{40+12+16.3}\)=0,12

c)nN₂=\(\frac{3,36}{22,4}\)=0,15

\(N_{Al}=0,25.6.10^{23}=1,5.10^{23}\)

\(n_{CO_2}=\dfrac{22}{44}=0,5mol\) 

\(\Rightarrow N_{CO_2}=0,5.6.10^{23}=3.10^{23}\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(\Rightarrow N_{H_2}=0,15.6.10^{23}=9.10^{22}\)

b. 

\(n_{N_2}=\dfrac{14}{28}=0,5\)

\(n_{Ca\left(NO_3\right)_2}=\dfrac{16,4}{164}=0,1mol\)

\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05mol\)

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) 

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ n_{CO_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

b) 

\(n_{Cl_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\Rightarrow m_{Cl_2}=n.M=0,06.71=4,26\left(mol\right)\\ n_{Na_2CO_3}=n.M=0,5.106=53\left(g\right)\)

c) 

\(V_{N_2\left(\text{đ}ktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{4,8}{32}=0,15\left(mol\right)\Rightarrow V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,15.22,4=3,36\left(l\right)\)

\(3.1.\left(a\right)M_P=31\left(g/mol\right);\\ M_{Fe}=56\left(g/mol\right);\\ M_{H_2}=2\left(g/mol\right);\\ M_{O_2}=32\left(g/mol\right)\\ \left(b\right).M_{P_2O_5}=31.2+16.5=142\left(g/mol\right);\\ M_{Fe_3O_4}=56.3+16.4=232\left(g/mol\right);\\ M_{HCl}=1+35,5=36,5\left(g/mol\right);\\ M_{BaO}=137+16=153\left(g/mol\right)\\ c.M_{H_2SO_4}=2+32+16.4=98\left(g/mol\right);\\ M_{ZnCl_2}=65+35,5.2=136\left(g/mol\right);\\ M_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342\left(g/mol\right);\\ M_{Ca\left(OH\right)_2}=40+17.2=74\left(g/mol\right)\)

\(3.2\left(a\right).n_{CH_4}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \left(b\right).n_{CuO}=\dfrac{2}{80}=0,025\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{3,42}{342}=0,01\left(mol\right)\)

\(n_{N_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\\ n_{Ca}=\dfrac{12.10^{23}}{6.10^{23}}=2\left(mol\right)\)

a) \(n_{Al\left(OH\right)_3}=\frac{7,8}{78}=0,1mol\)

b) \(n_{Al_2\left(SO_4\right)_3}=\frac{34,2}{342}=0,1mol\)

c) \(n_{H_2}=\frac{3,36}{22,4}=0,15mol\)