What is the Mass of LiHCO_3 in the Original Mixture?

[PhizKid]

Homework Statement

Given the reaction: $$2LiHCO_3 + SiO_2 \rightarrow Li_{2}CO_3 + SiO_2 + CO_2 + H_{2}O$$ where $$SiO_2$$ is unaffected, the mass of the $$2LiHCO_3 + SiO_2$$ is 9.62 g. and $$Li_{2}CO_3 + SiO_2$$ is 6.85 g., find:

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

The Attempt at a Solution

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

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

c) Since the mass of $$SiO_2$$ does not change, 9.62 g. in the original - 0.79 g. of $$LiHCO_3$$ = 8.83 g. $$SiO_2$$. But 8.83 g. of $$SiO_2$$ + 2.77 g. of $$CO_2 + H_{2}O$$ 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

 

[trollcast]

Don't trust my method on this one but I can get the numbers to add up.

Take your mass of C02 + H20 and work out the number of moles of CO2 AND H2O molecules you have.

Then use the ratios of the balancing numbers to work out the moles of the other reactants.

Then multiply by the RFM of the reactant to get the mass.

Ok that probably makes very little sense, but it might help.

 

[Borek]

So if 2.77 g. are lost to $$CO_2 + H_{2}O$$, then there was initially 0.79 g. of $$LiHCO_3$$

Check your math. Logic is sound, but you got your proportions wrong.