[daigo1]

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

[DrCMS]

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.