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Balancing questions » determining composition

» Equation balancing and stoichiometry calculator.

Solid NaOH is very efficient in absorbing CO2 from the air. I have a jar of old NaOH pellets contaminated with Na2CO3. To check their composition I have dissolved 0.949 g of solid in 100 mL of water. 35.4 mL of 0.650 M HCl was needed for complete neutralization. What is the percentage composition of pellets?

Such questions require us to set up a system of simultaneous equations to be solved. We know that sample contains two substances - let's say it contains mNaOH grams of NaOH and mNa2CO3 grams of sodium carbonate. We know total mass of the sample - and it can be expressed as sum of massses of both substances:

mNaOH + mNa2CO3 = 0.949 g

Preparation of second equation requires some knowledge about reactions taking place during neutralization. There are two such reactions, one for NaOH:

NaOH + HCl -> NaCl + H2O

and one for Na2CO3:

Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O

What is important here is molar ratio of reacting substances. From the first equation we see that NaOH reacts with HCl in 1:1 molar ratio, so to neutralize nNaOH moles of NaOH we need the same number of moles of HCl. In the case of Na2CO3 we need twice more HCl. For calculations we will also need molar masses of both substances, as in this case it is number of moles what counts. Note, that volume of the solution prepared (100 mL) is of no use here. To convert mass of the substance to amount of moles it is enough to divide it by molar mass, so we can write second equation in the form:

mNaOH/40.00 + 2mNa2CO3/106.0 = nHCl

where nHCL is number of moles of HCl used in the neutralization, 40.00 g is a molar mass of NaOH and 106.0 g is a molar mass of Na2CO3. To calculate nHCL we will use definition of molar concentration, rearranged to form

n = CV = 0.0354 L × 0.650 M = 0.0230 mole

and finally our set of equations looks this way:

mNaOH + mNa2CO3 = 0.949 g

mNaOH/40.00 + 2mNa2CO3/106.0 = nHCl

It is not that difficult to solve, and the values calculated are:

mNaOH = 0.830 g

mNa2CO3 = 0.119 g

Which gives percentage composition of

0.830/0.949 × 100% = 87.5% NaOH

and

0.119/0.949 × 100% = 12.5% Na2CO3

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