# How do you write the chemical equation (with state symbols) for the reaction? Which of the two reactants is the limiting reagent? How do you calculate the mass of carbon dioxide produced and the mass of the excess reactant left unreacted?

## 5.00 g of solid copper (II) carbonate was allowed to react with 1.17 g of aqueous hydrochloric acid, to give aqueous copper (II) chloride, water and carbon dioxide gas.

Oct 11, 2016

The stoichiometric equation is:

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

#### Explanation:

How could I write this? Well, I simply learned that that metal carbonates reacted with hydrochloric acids to form the metal chloride and carbon dioxide and water:

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

These equations simply have to be learned. But you are helped by the fact that mass and charge are always balanced. Are they balanced here? Are you sure?

If we dig a little deeper, what we have labelled as $M C {l}_{2} \left(a q\right)$ is probably the hexa-aqua complex, ${\left[M {\left(O {H}_{2}\right)}_{6}\right]}^{2 +}$ and $2 \times C {l}^{-} \left(a q\right)$, but the given equation allows us to make stoichiometric predictions.

$\text{Moles of copper carbonate}$ $=$ $\frac{5.00 \cdot g}{123.56 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0405 \cdot m o l$.

Now if the hydrochloric acid were prepared with $1.17 \cdot g$ of hydrogen chloride gas (this is not what the question proposed; it could be $1.17 \cdot g$ of conc. hydrochloric acid), there are $\frac{1.17 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0321 \cdot m o l$, and clearly, the hydrochloric acid is the limiting reagent.

And thus only $0.0161 \cdot m o l$ of $C u C {l}_{2} \left(a q\right)$ and $C {O}_{2} \left(g\right)$ would be produced.

Anyway, I will leave it to you to clarify the question.