Thread: [HELP!] Chemistry [FAST]

Barium nitrate and sodium carbonate react to form barium carbonate and sodium nitrate.

If 234 grams of barium nitrate and 269 grams of sodium carbonate are reacted, which is the limiting reactant?

How many grams of barium carbonate precipitant could be produced? (Show work Please!)

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both are salts so we assume that in the reaction the salts will switch cations (or anions if you want to say that).
Ba(NO3)2 = 261.37 g/mol
Na2CO3 = 84.9947 g/mol
CO3 = -2 charge
NO3 = -1 charge
Ba(NO3)2 + Na2CO3 => BaCO3 + 2Na(NO3)

since its a 1 to 1 ratio, and you start with similar-ish amounts of substance, because the density of Ba(NO3)2 is so much higher, we can assume that there are less moles of it, and therefore that it is the limiting reactant.

Check:
234/ 261.37 moles of barium nitrate and 269/ 84.9947 moles sodium nitrate
~.9 and ~ 3.1 (rough guesses)
.9 < 3.1; Therefore our assumption was valid.