# What volume of carbon dioxide gas, CO 2 , would be produced at STP if 23.40 g of aluminum carbonate reacts with excess hydrochloric acid?

## Al 2 (CO 3 ) 3 (aq) + 6HCl (aq) ----> 2AlCl 3 (aq) + 3H 2 O (l) + 3CO 2 (g) What volume of carbon dioxide gas, CO 2 , would be produced at STP if 23.40 g of aluminum carbonate reacts with excess hydrochloric acid? a) 2.24 L b) 6.72 L c) 22.4 L d) 672 L

Jun 18, 2018

Would it not be $\left(b\right)$?

#### Explanation:

We gots the stoichiometric equation..

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

$\text{Moles of aluminum salt} = \frac{23.4 \cdot g}{233.99 \cdot g \cdot m o {l}^{-} 1} = 0.100 \cdot m o l$

And so we should generate a $0.300 \cdot m o l$ quantity of gas. Your syllabus might specify that a mole of Ideal Gas occupies $24.6 \cdot {\mathrm{dm}}^{3}$ at $298 \cdot K$ and $0.987 \cdot a t m$..of course I do not know what it specifies so it might be an idea to find out...

And so we take the product...$0.300 \cdot m o l \times 24.6 \cdot L \cdot m o {l}^{-} 1 = 7.4 \cdot L$...this is close enuff to $b$...clearly we used a molar volume different from their specification...