PROBLEM 37P

Consider the balanced equation for the combustion of methane, a component of natural gas:

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

Complete the table with the appropriate masses of reactants and products. If the mass of a reactant is provided, fill in the mass of other reactants required to completely react with the given mass, as well as the mass of each product formed. If the mass of a product is provided, fill in the required masses of each reactant to make that amount of product, as well as the mass of the other product that forms.

Mass CH4 |
Mass O2 |
Mass CO2 |
Mass H2O |

____ |
2.75 kg |
____ |
____ |

22.32 g |
____ |
____ |
____ |

____ |
____ |
____ |
11.32 g |

____ |
____ |
2.94 g |
____ |

3.18 kg |
____ |
____ |
____ |

____ |
____ |
2.35 × 103 kg |
____ |

Solution 37P

The completed table is as given below

Mass CH4 | Mass O2 | Mass CO2 | Mass H2O |

0.687 Kg | 2.75 kg | 1.89Kg | 0.562Kg |

22.32 g | 89.28 g | 61.38g | 50.22g |

5.03g | 20.12g | 13.83g | 11.32 g |

1.069g | 4.41g | 2.94 g | 2.40g |

3.18 kg | 12.72kg | 8.74kg | 7.15Kg |

0.854 × 103 kg | 3.41 × 103 kg | 2.35 × 103 kg | 1.92× 103 kg |

Steps for calculation

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

We know from the equation that 1 mole CH4 combines with 2 moles of O2 to give 1 mole CO2 and 2 moles of H2O

Thus 16 gms of CH4combines with 64 gms of O2 to give 44 gms CO2 and 36gms of H2O

1 st row calculation

Mass of CH4=2.75103g=0.687kg

Mass of CO2=2.75103g=1.89kg

Mass of H2O=2.75103g=0.562kg

2nd row calculation

Mass of O2=22.32g=89.28g

Mass of CO2=22.32g=61.38g

Mass of H2O=22.32g=50.22g

3rd...