1. Lithium hydrogen carbonate decomposes by heat; the reaction is represented by

Question

1. Lithium hydrogen carbonate decomposes by heat; the reaction is represented by the balanced equation:

2LiHCO3 (s) --------> Li2CO3(s) + CO2(g) + H20 (g)

a.) If 0.24 mol of LiHCO3 is decomposed completely, how many mol of Li2CO3 will be formed?

b.) If 136 g of LiHCO3 are decomposed completely, how many g of Li2CO3 will be formed?

c.) What is the combined weight of the CO2 and H20 in b?

2.

a.) A sample of LiHCO3 and sand mixture weighing 9.62 g is decomposed by heat. After cooling, the residue, which is composed of Li2CO3 and sand, is formed to weigh 6.85 g. Assuming that the sand in the sample is unaffected, calculate the weight loss due to the evolution of gaseous CO2 and H2O.

b.) Calculate the weight of LiHC03 in the original mixture.

c.) Calculate the weight of sand in the mixture.

d.) Calculate the percent of lithium hydrogen carbonate in the mixture.

e.) What must be the percent of sand in the mixture?

Explanation / Answer

1a ) 0.24/2 = 0.12 mole Li2CO3 is formed

b) moles of LiHCO3 = 136/67.96 = 2 , moles of L2CO3 = 2/2 = 1

mass of Li2CO3 = 1 x 73.89   = 73.89 gm

c) combined wt = ( 1 x 18) +(1x44) = 62 gm

2) a) weight loss =(9.62-6.85) = 2.77 gm ,

b) loss weight = 2.77 gm conists CO2 and H2O

combined mol wt = ( 44+18) = 62 gm per 1 mole

hence moles of gas lost = ( 2.77/62) = 0.04468,

moles of LiHCO3 = 2 x 0.04468 = 0.089355 ,

wt of LiHCO3 = 67.96 x 0.089355 = 6.073 gm ,

c) wt of sand = 9.62-6.073 =3.547 gm,

d) % LiHCO3 = ( 6.073/9.62) x100 = 63.13 %

e) % sand = 100-63.13 = 36.87 %

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