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If we use the reaction Sulphuric Acid with Sodium Hydroxide. The reaction equation is

H2SO4 + 2NaOH = Na2SO4 + 2H2O

Note the molar ratios are 1:2::1:2

However if we have say 75 grams sulphuric acid. what masses are required by sodium hydroxide to make the reaction go to completion. Also what masses of sodium sulphate and water will produce.

First we calculate the moles of H2SO4 . To do this we find the atomic masses.

H x 2 = 1 x 2 = 2

S x 1 = 32 x 1 = 32

O x 4 = 16 x 4 = 64

2 + 32 + 64 = 98

So moles(H2SO4) = 75 / 98 = 0.7653moles ( equivalent to one(1) ratio).

Hence moles (NaOH) = 0.7653 x 2 = 1.5306 ( equivalent to two(2) ratios)

Hence masses of NaOH required for complete reaction is

mass(g) = 1.5306 x (23 + 16 + 1) = 61.44 g.

So total mass of reactants is 75 + 61.44 = 136.44. g

As per the Law of conservation of Mass , the total mass of the products will be 136.44 g. This figure has to be apportioined for the product masses.

Since there are two moles of water produced , then mass of water is

Mass(H2O) =1.5306 x 18) = 27.5508 g

Hence mass of sodium sulphate is 136.44 - 27.5508n = 108.892 g

To verfiy this

First calculate the Relative Molecular mass of Na2SO4

which is

Na x 2 = 23 x 2 = 46

S x 1 = 32 x 1 = 32

O x 4 = 16 x 4 = 64

46 + 32 + 64 = 142

Mass(Na2SO4) = 0.7653 x 142 = 108.67 g ~ 108.892g

The difference in these figures is due to estimations, and calculator errors.

Hopefully you can see ,what stoichiometry is all about. !!!!!

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lenpollock

Lvl 16
2y ago

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