Question

Tetraboric acid, H2B4O7, is prepared by heating boric acid, H3BO3 (a condensation reaction involving water loss). If 400.0 mmol H3BO3 are used, what mass (g) of H2O is formed, assuming quantitative stoichiometric conversion?

I know the answer is 9.01 but how do you get the condensation reaction?

Answer 1

Tetraboric acid, H2B4O7, is prepared (means product) by heating boric acid, H3BO3 (means reactant), involving water loss (means it is also one of the product) loss from reactant H3BO3.

First we have to write the balanced chemical reaction equation to find out exact stoichiometrics

Hey first we have to write the basic equation

H₃BO₃ =======> H₂B₄O₇ + H₂O

now we have to find out the coefficients, this needs a lot of experience, balance boron first in product side we found 4 boron atoms ( H₂B₄O₇) then go for H₃BO₃ reactant side put 4 as coefficient, check whether all atoms are balanced, yes all are balanced now, thats it. (OR you may use any software also for this purpose). I just ended-up with following one.

4 H₃BO₃ =======> H₂B₄O₇ + 5 H₂O

according to above equation 4 moles H₃BO₃ gives 5 moles H₂O

400.0 mmol H₃BO₃ = 0.4 mol H₃BO₃            --------------------     since 1 mol = 1000 mmol or 1mmol = 0.001 mol

4 mol H₃BO₃ -------   5 mol H₂O

0.4 mol H₃BO₃   ---- 0.5 mol H₂O     (Just simply we divided both by 10) as we have 0.4 mol H3BO3

convert moles H₂O into mass H₂O

mass of H₂O = moles of H₂O x molar mass of H₂O

mass of H₂O = 0.5 mol x 18.01528 g/mol

mass of H₂O = 9.00764 ~ 9.01 grams

mass (g) of H₂O is formed = 9.01 grams.

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