# If 10 mol of CaC_2 reacts with H_2O, how many moles of C_2H_2 will be produced?

## $C a {C}_{2} + 2 {H}_{2} O \to C a {\left(O H\right)}_{2} + {C}_{2} {H}_{2}$

Jun 1, 2017

This is an acid base reaction.............

#### Explanation:

With water......

$C {a}^{2 +} {\left\{C \equiv C\right\}}^{2 -} + 2 {H}_{2} O \rightarrow C a {\left(O H\right)}_{2} + H C \equiv C H \left(g\right) \uparrow$

And the stoichiometry clearly indicates that ONE mole of calcium carbide, a mass of $64.1 \cdot g$ reacts with TWO moles of water, $36 \cdot g$ to give $74.1 \cdot g$ calcium hydroxide, and $26 \cdot g$ acetylene.

Is mass conserved in this reaction? Should it be?

And if you decompose $10 \cdot m o l$ calcium carbide you clearly gets $10 \cdot m o l$ acetylene. Agreed?

Jun 1, 2017

$\text{10 mols}$ of ${\text{C"_2"H}}_{2}$.

Since you already have the chemical reaction balanced, I think you're past the hard part. Now we can use the chemical reaction. Reading off the coefficients, we have the stoichiometric ratios at our disposal.

For example...

• For every one mol of ${\text{CaC}}_{2}$, one mol of "Ca"("OH")_2 is produced.

• For every one mol of ${\text{CaC}}_{2}$, one mol of ${\text{C"_2"H}}_{2}$ is produced.

• For every one mol of ${\text{CaC}}_{2}$, two mols of water react.

... and so on. You didn't say how much water reacts, so we assume excess water.

From the second bullet point, we know that ${\text{CaC}}_{2}$ and ${\text{C"_2"H}}_{2}$ have a $1 : 1$ mol ratio. For emphasis, this is saying:

$\textcolor{red}{1} C a {C}_{2} + 2 {H}_{2} O \to C a {\left(O H\right)}_{2} + \textcolor{red}{1} {C}_{2} {H}_{2}$

So, you apparently already have the number written down. It is $\textcolor{b l u e}{\text{10 mols}}$ of ${\text{C"_2"H}}_{2}$ that is produced (assuming the water is in excess).