What volume of gas is generated at STP when 4.6 g of Aluminum Carbonate is placed in 0.75 M Nitric acid?

Mar 4, 2015

You'd produce $\text{1.3 L}$ of $C {O}_{2}$ gas at STP with that much aluminium carbonate.

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.

$A {l}_{2} {\left(C {O}_{3}\right)}_{3 \left(a q\right)} + 6 H N {O}_{3} \left(a q\right) \to 2 A l {\left(N {O}_{3}\right)}_{3 \left(a q\right)} + 3 {H}_{2} {O}_{\left(l\right)} + 3 C {O}_{2 \left(g\right)}$

Notice that you have a $\text{1:3}$ mole ratio between aluminium carbonate and carbon dioxide, i.e. you'll produce three times as many moles of the latter than the number of moles of the former that react.

The number of moles of aluminium carbonate is

$\text{4.6 g" * "1 mole"/"234.0 g" = "0.0197 moles}$

Now, a 1.0-L sample of nitric acid will have

$C = \frac{n}{V} \implies n = C \cdot V = \text{0.75 M" * "1.0 L" = "0.75 moles nitric acid}$

which is more than enough to make sure that the nitric acid is not limiting the reaction.

According to the $\text{1:6}$ mole ratio that exists between aluminium carbonate and nitric acid, you would need a minimum of

$\text{0.0197 moles aluminijm carbonate" * "6 moles nitric acid"/"1 mole aluminium carbonate" = "0.118 moles nitric acid}$

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

$\text{0.0197 moles aluminium carbonate" * "3 moles carbon dioxide"/"1 mole aluminium carbonate" = "0.0591 moles}$

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

${n}_{C {O}_{2}} = \frac{V}{V} _ \left(\text{molar") => V = n_(CO_2) * V_("molar}\right)$

$V = \text{0.0591 moles" * "22.4 L"/"1 mole" = "1.32 L}$

Rounded to two sig figs, the number of sig figs in both 4.6 g and in 0.75 M, the answer will be

$V = \text{1.3L}$

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.