Question

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

Answer 1

Would it not be `(b)`?

Explanation:

We gots the stoichiometric equation..

`Al_2(CO_3)_3(s) + 6HCl(aq) rarr 2AlCl_3(aq)+3H_2O(l) + 3CO_2(g)uarr`

`"Moles of aluminum salt"=(23.4*g)/(233.99 *g*mol^-1)=0.100*mol`

And so we should generate a `0.300*mol` quantity of gas. Your syllabus might specify that a mole of Ideal Gas occupies `24.6*dm^3` at `298*K` and `0.987*atm`..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*molxx24.6*L*mol^-1=7.4*L`...this is close enuff to `b`...clearly we used a molar volume different from their specification...