PROBLEM 38P

Consider the balanced equation for the combustion of butane, a fuel often used in lighters:

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

Complete the table showing 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 CH4H10 |
Mass O2 |
Mass CO2 |
Mass H2O |

____ |
1.1 g |
____ |
____ |

5.22 g |
____ |
____ |
____ |

____ |
____ |
10.12 g |
____ |

____ |
____ |
____ |
9.04 g |

232 mg |
____ |
____ |
____ |

____ |
____ |
118 mg |
____ |

Solution 38P

The completed table is as given below

Mass C4H10 |
Mass O2 |
Mass CO2 |
Mass H2O |

0.306g |
1.1 g |
0.930g |
0.475g |

5.22 g |
18.72g |
15.84g |
8.1g |

3.33g |
11.96g |
10.12 g |
5.175g |

5.82g |
20.89g |
17.67g |
9.04 g |

232 mg |
832mg |
704mg |
360mg |

38.8mg |
139.4mg |
118 mg |
60.34mg |

Steps for calculation

2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g)

We know from the equation that 2 mole C4H10 combines with 13moles of O2 to give 8 moles CO2 and 10 moles of H2O

Thus 116 gms of C4H10combines with 416 gms of O2 to give 352 gms CO2 and 180gms of H2O

1 st row calculation

Mass of C4H10=1.1g=0.306g

Mass of CO2=1.1g=0.930kg

Mass of H2O=1.1g=0.475kg

2nd row calculation

Mass of O2=5.22g=18.72g

Mass of CO2=5.22g=15.84g

Mass of H2O=5.22g=8.1g

3rd row calculation

Mass of C4H10=10.12g=3.33g

Mass of O2=10.12g=11.96g

Mass of CO2=10.12g=5.175 g

4th row calculation

Mass of C4H10=9.04g=5.82g

Mass of O2=9.04g=20.89g

Mass of H2O=9.04g=17.67g

5th row calculation

Mass of O2=232mg=832 mg

Mass of CO2=232mg=704mg

Mass of H2O=232mg=360mg

6th row calculation

Mass of C4H10=118mg=38.8mg

Mass of O2=118mg=139.45mg

Mass of H2O=118mg=60.34mg