May 9th, 2018, 01:17 AM |
#1 |

Member Joined: May 2015 From: Australia Posts: 77 Thanks: 7 | Stoichiometry
Find the mass of iron (III) chloride that can be formed by burning 14g of iron in excess chlorine. 2Fe + 3Cl2 --> 2FeCl3 My answer: m(Fe)=14g Mr(Fe)=55.84g/mol n(Fe)=14g/(55.84g/mol) =0.251mol Ratio Fe:FeCl3 2:2 Thus, n(FeCl3) = 0.251mol m(FeCl3) = 0.251mol x 162.147g/mol = 40.7g My concern is that I might have calculated the Molecular Mass (Mr) incorrectly. Would the molecular mass of 2Fe be 55.84g/mol or 111.68g/mol Similarly, would the molecular mass of 2FeCl3 be 162.147g/mol or 324.38g/mol. |

May 9th, 2018, 02:27 AM |
#2 |

Senior Member Joined: Jun 2015 From: England Posts: 830 Thanks: 244 |
It's the atomic mass for the iron and the molecular mass for the chlorine. But each is multiplied by the coefficient counting the number of atoms or molecules So the coefficients are 2 for the iron and 3 for the chlorine. Remembering that the molecular mass of chlorine = 2 x atomic mass of chlorine. |

May 9th, 2018, 02:36 AM |
#3 |

Member Joined: May 2015 From: Australia Posts: 77 Thanks: 7 |
So would this be the answer (taking into account the coefficient when calculating molecular mass)? Find the mass of iron (III) chloride that can be formed by burning 14g of iron in excess chlorine. 2Fe + 3Cl2 --> 2FeCl3 My answer: m(Fe)=14g Mr(Fe)=55.84g/mol * 2 = 111.68 g/mol n(Fe)=14g/(111.68g/mol) =0.125mol Ratio Fe:FeCl3 2:2 Thus, n(FeCl3) = 0.125mol m(FeCl3) = 0.125mol x 324.38g/mol = 40.5g |

May 9th, 2018, 04:29 AM |
#4 |

Senior Member Joined: Jun 2015 From: England Posts: 830 Thanks: 244 |
I think you are making this very complicated. For every atom of iron there are 3 atoms of chlorine in the ferric chloride. The atomic mass of iron is 55.8 so and of chlorine is 35.5. So 55.5 grams of iron produces {55.5 + 3(35.5)} = 162 grams of ferric chloride. So 1 gram of iron produces $\displaystyle \frac{162}{{55.5}}$ grams of ferric chloride So 14 grams of iron produces $\displaystyle 14*\frac{{162}}{{55.5}} = 40.86$ g of ferrric chloride |

May 9th, 2018, 09:54 AM |
#5 | |

Math Team Joined: May 2013 From: The Astral plane Posts: 1,810 Thanks: 724 Math Focus: Wibbly wobbly timey-wimey stuff. | Quote:
$\displaystyle \frac{14~g~Fe}{1} \cdot \frac{1~mol~Fe}{55.845~g~Fe} \cdot \frac{2~mol~FeCl_3}{2~mol~Fe} \cdot \frac{162.195~g~FeCl_3}{1~mol~FeCl_3} = 40.66~g~FeCl_3$ The second to last factor comes from comparing the number of mols of each substance in the chemical reaction. Note how the different units cancel. -Dan | |

May 9th, 2018, 10:50 AM |
#6 |

Senior Member Joined: Jun 2015 From: England Posts: 830 Thanks: 244 |
Thank you Dan for correcting my arithmetic. My point is that you don't need the reaction equation at all, just the formula for ferric chloride. The question says in "excess chlorine" so all the iron is used up. |

May 9th, 2018, 04:33 PM |
#7 |

Math Team Joined: May 2013 From: The Astral plane Posts: 1,810 Thanks: 724 Math Focus: Wibbly wobbly timey-wimey stuff. | |