There are two unknowns in this problem - the mass of potassium carbonates and the mass of sodium carbonate. Let's designate the grams of potassium carbonate as our first unknown (you may want to call it gKcarb, or x, some other variable name that makes sense to you) and the grams of sodium carbonate as our second unknown(you may want to call it gNacarb, or y, some other variable name that makes sense to you). Set up an equation for the sum of your two unknowns. Starting with 'unknown' grams of potassium carbonate, use stoichiometry to calculate the number of moles of nitric acid that would react with the potassium carbonate. Your answer will have a variable for your unknown grams of potassium carbonate in it. Starting with 'unknown' grams of sodium carbonate, use stoichiometry to calculate the number of moles of nitric acid that would react with the sodium carbonate. Your answer will have a variable for your unknown grams of sodium carbonate in it. Set up an equation for what you get if you add these two quantities.

Respuesta :

This problem, is describing two scenarios, the first one, a reaction between potassium carbonate and nitric acid, and also this very same acid with sodium carbonate. In addition, it is asked to set up two equations whereby you can get the grams of nitric acid needed to react with the unknown grams of both carbonates.

In this case, we can start off by designating the unknown mass of potassium carbonate as X and that of sodium carbonate as Y, so that we will be able to provide a reliable answer. Next, we write the corresponding chemical equations that take place:

K2CO3 + 2HNO3 --> 2KNO3 + H2O + CO2

Na2CO3 + 2HNO3 --> 2NaNO3 + H2O + CO2

After that, we can set up the conversion by considering the following calcultion track:

g Carbonate --> mol Carbonate --> mol HNO3 --> g HNO3

The conversion from grams to moles involve the carbonates' molar mass and the conversion from moles of nitric acid to grams, its molar mass well. In addition, we need the 1:2 mole ratio of the carbonates to nitric acid that it is evidenced in the reaction.

Therefore, the resulting equations that can be set up are shown as follows:

[tex]X g K_2CO_3 *\frac{1mol K_2CO_3 }{138.2 gK_2CO_3 } \frac{2molHNO_3}{1molK_2CO_3 } *\frac{63.1gHNO_3}{1molHNO_3} \\\\Y g Na_2CO_3 *\frac{1mol K_2CO_3 }{105.99 gK_2CO_3 } \frac{2molHNO_3}{1molNa_2CO_3 } *\frac{63.1gHNO_3}{1molHNO_3}[/tex]

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