Bài 1:
\(M_{NH_4NO_3}=14+1.4+14+3.16=80\)
% m N = \(\dfrac{2.14}{80}.100\%=35\%\)
\(M_{\left(NH_4\right)_2SO_4}=2.\left(14+1.4\right)+32+4.16=132\)
% m N = \(\dfrac{2.14}{132}.100\%=21,21\%\)
\(M_{\left(NH_2\right)_2CO}=2.\left(14+2\right)+12+16=60\)
% m N = \(\dfrac{2.14}{60}.100\%=46,7\%\)
\(M_{NH_4Cl}=14+1.4+35,5=53,5\)
% m N = \(\dfrac{28}{53,5}.100\%=52,3\%\)
Bài 2:
a) \(n_{SO_2}=\dfrac{5,6}{64}=0,0875\left(mol\right)\)
\(n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)\)
nhỗn hợp = 0,0875 + 0,25 = 0,3375 (mol)
b) mhỗn hợp = 0,3375.(64 + 32) = 32,4 (g)
Bài 3:
\(n_P=\dfrac{15,5}{31}=0,5\left(mol\right)\)
PT: 4P + 5O2 → 2P2O5
mol 0,5 → 0,625 0,25
\(V_{O_2\left(đktc\right)}=0,625.22,4=14\left(l\right)\)
Vkhông khí = 14 : 20% = 70 (l)