If 88 g of NaCl reacted with an unknown amount of Ca(OH)2 and 95 g of NaOH was produced with 170 g of CaCl2, how many grams of Ca(OH)2 was reacted?

Answer 1

In the real world, this reaction would not take place, i.e. NO REACTION. The reason is that all reactants (and potential products) are aqueous, and there is no precipitate or gas or liquid formed. Even though Ca(OH)₂ is partially soluble, the ionic equation would be:

2Na⁺(aq) + 2Cl^-(aq) + Ca²⁺(aq) + 2OH^-(aq) => 2Na⁺(aq) + 2OH^-(aq) + Ca²⁺(aq) + 2Cl^-(aq)

This may have been a question to see if you understand the concept of CONSERVATION OF MASS. This law tells us that the mass of the reactants must equal the mass of the products, as we can neither create nor destroy matter. So, if that is the intent of the question, then we can look at the problem as follows:

88 g NaCl + x g Ca(OH)₂ ==> 95 g NaOH + 170 g CaCl₂

mass of reactants = 88 g + x g

mass of products = 95 g + 170 g = 265 g

88 g + x g = 265 g

x g = 177 g of Ca(OH)₂ would have reacted