\(3.a.M_{CaCO_3}=40+12+16.3=100\left(\dfrac{g}{mol}\right)\)
\(b.n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)
\(c.d_{\dfrac{C_2H_4}{SO_2}}=\dfrac{28}{64}=0,4375\)
\(4.m_{hhA}=m_{N_2}+m_{CO_2}+m_{CH_4}=0,25.28+0,75.44+0,5.16=7+33+8=48\left(g\right)\)
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