Question

Answer 1

BT2:

\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

                                            0,5--->1

=> m = 1.18 = 18 (g)

BT3:

\(n_{CO_2}=\dfrac{22,4}{22.4}=1\left(mol\right)\)

PTHH: C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

                                                1----->1

=> m = 1.18 = 18 (g)

Answer 2

BT2:

\(n_{CH_4}=\dfrac{16}{16}=1\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(m=mH_2O=1.2.18=36\left(g\right)\)

BT3:

\(nC_2H_4=\dfrac{14}{28}=0,5\left(mol\right)\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(m=mH_2O=0,5.2.18=18\left(g\right)\)

Answer 3

nCO2=22,422.4=1(mol)nCO2=22,422.4=1(mol)

PTHH: C₂H₄ + 3O₂ --t^o--> 2CO₂ + 2H₂O

                                                1----->1

=> m = 1.18 = 18 (g)