What volume of gas is generated at STP when 4.6 g of Aluminum Carbonate is placed in 0.75 M Nitric acid?
SIDE NOTE. Since you didn't provide a volume for the nitric acid, I'll assume a 1-L sample; I've based my assumption on the fact that nitric acid will not act as a limiting reagent, i.e. you'll have excess nitric acid.
So, start with the balanced chemical equation
Notice that you have a
The number of moles of aluminium carbonate is
Now, a 1.0-L sample of nitric acid will have
which is more than enough to make sure that the nitric acid is not limiting the reaction.
According to the
In this case, you indeed have more nitric acid than you need.
Use their respective mole ratio to determine how many moles of carbon dioxide will be produced
At STP (273,15 K and 1 atm), 1 mole of any ideal gas occupies exactly 22.4 L - this is known as the molar volume of a gas at STP.
Therefore, the volume of gas produced will be
Rounded to two sig figs, the number of sig figs in both 4.6 g and in 0.75 M, the answer will be
SIDE NOTE If a volume for the nitric acid comes up, determine the number of moles of nitric acid you'd have in that particular volume and compare it to the minium number of moles required for that much aluminium carbonate - if the number's smaller, the nitric acid will be a limiting reagent and you'd have to recalculate the moles of aluminium carbonate that react.