Balancing & stoichiometry lectures

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Balancing questions » limiting reagent

In 1930 workers building the house of my Grandparents prepared slaked lime by adding 300 L of water to 1000 kg of burnt lime. Grandpa argued that there wasn't enough water, but they didn't want to hear. Who was right?

The first thing to do is to write a balanced reaction equation (in a way all stoichiometric problems are boring - solutions always start the same way):

CaO + H2O -> Ca(OH)2

We know mass of lime, but we don't know mass of water - however, that's a thing everyone should know. 1 L of water weights almost 1 kg. The rest is a simple limiting reagent problem, that we will tackle calculating the mass of water needed to fully react with CaO:

 CaO + H2O -> Ca(OH)2 56.08 g : 18.02 g

Oops... seems we may have a problem - molar masses are given in grams, while masses of reagents are given in kg... But does it really matter? 56.08 g of CaO reacts with 18.02 g H2O. We can say as well that 56.08 stones of CaO reacts with 18.02 stones of H2O - the mass unit used doesn't change mass ratio, so we can safely write:

 CaO + H2O -> Ca(OH)2 56.08 g : 18.02 g 1000 kg : x kg

So proportion is

56:08 g : 18.02 g = 1000 kg : x kg

and x = 56.08 g / 18.02 g × 1000 kg = 321.3 kg

In our final formula for x grams cancel out and final unit of our result is kg, that confirms we are right using different units in ratios.

So, going back to the question - Grandpa was right. Workers were about 2 buckets of water short (in fact they should use even more water, as slacked lime must be wet when used).