January 26, 2021, 01:35:14 PM
Forum Rules: Read This Before Posting


Topic: Weight loss by decomposition  (Read 1066 times)

0 Members and 1 Guest are viewing this topic.

Offline daigo1

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-1
Weight loss by decomposition
« on: March 02, 2013, 11:20:34 AM »
Given the reaction: [tex]2LiHCO_3 + SiO_2 \rightarrow Li_{2}CO_3 + SiO_2 + CO_2 + H_{2}O[/tex] where [tex]SiO_2[/tex] is unaffected, the mass of the [tex]2LiHCO_3 + SiO_2[/tex] is 9.62 g. and [tex]Li_{2}CO_3 + SiO_2[/tex] is 6.85 g., find:

a) mass loss due to [tex]CO_2 + H_{2}O[/tex]
b) mass of [tex]LiHCO_3[/tex] in the original mixture
c) mass of [tex]SiO_2[/tex] in the new mixture



a)The mass lost is just 9.62 g. - 6.85 g. = 2.77 g. of [tex]CO_2 + H_{2}O[/tex]

b) For every 136 g. of [tex]LiHCO_3[/tex], 62 g. are lost to [tex]CO_2 + H_{2}O[/tex]. So if 2.77 g. are lost to [tex]CO_2 + H_{2}O[/tex], then there was initially 0.79 g. of [tex]LiHCO_3[/tex]

c) Since the mass of [tex]SiO_2[/tex] does not change, 9.62 g. in the original - 0.79 g. of [tex]LiHCO_3[/tex] = 8.83 g. [tex]SiO_2[/tex]. But 8.83 g. of [tex]SiO_2[/tex] + 2.77 g. of [tex]CO_2 + H_{2}O[/tex] is greater than 9.62 g. of the original mixture, so mass is not conserved. I don't think this is possible so I don't know what to do

Offline DrCMS

  • Chemist
  • Sr. Member
  • *
  • Posts: 1268
  • Mole Snacks: +207/-81
  • Gender: Male
Re: Weight loss by decomposition
« Reply #1 on: March 02, 2013, 02:04:28 PM »

b) For every 136 g. of [tex]LiHCO_3[/tex], 62 g. are lost to [tex]CO_2 + H_{2}O[/tex]. So if 2.77 g. are lost to [tex]CO_2 + H_{2}O[/tex], then there was initially 0.79 g. of [tex]LiHCO_3[/tex]

Try doing your maths again.

Sponsored Links