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.