# In the reaction CaC_2(s) + 2H_2O(l) -> C_2H_2(g) + Ca(OH)_2(aq), if 23 g of CaC_2 are consumed in this reaction, how much H_2O is needed?

Feb 26, 2016

Approx. $13$ $m L$ of water.

#### Explanation:

Moles of calcium carbide: $\frac{23 \cdot g}{64.10 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.359$ $m o l$.

Given the equation, $\text{2 equiv}$ of water are required:

$0.359 \times 2 \cdot m o l$ water, $=$ $0.718$ $m o l$ water.

$\text{Equivalent mass}$ $=$ $0.718 \cdot m o l \times 18.01 \cdot g \cdot m o {l}^{-} 1$ $=$ $\text{approx. 12.9 g}$ water.

Feb 26, 2016

You will need $\text{13 g H"_2"O}$.

#### Explanation:

Balanced Equation
$\text{CaC"_2("s")+"2H"_2"O(l)}$$\rightarrow$$\text{C"_2"H"_2("g")+"Ca(OH)"_2("aq")}$

We will make the following conversions to answer this question.

${\text{mass CaC}}_{2}$$\rightarrow$${\text{mol CaC}}_{2}$$\rightarrow$$\text{mol H"_2"O}$$\rightarrow$$\text{mass H"_2"O}$

In order to do this, we need to determine the molar mass of each compound. We can calculate it or look it up on a reputable source. I like to use The PubChem Project.

Molar Masses

$\text{CaCl"_2} :$ $\text{64.0994 g/mol}$
https://pubchem.ncbi.nlm.nih.gov/compound/6352section=Top

$\text{H"_2"O} :$ $\text{18.01528 g/mol}$
https://pubchem.ncbi.nlm.nih.gov/compound/962section=Top

We also need the mole ratio between the two compounds.

From the balanced equation we can see that the mole ratio between $\text{CaC"_2}$ and $\text{H"_2"O}$ is $\text{1 mol CaC"_2: "2 mol H"_2"O}$.

Solution

Convert ${\text{mass CaC}}_{2}$$\rightarrow$${\text{mol CaC}}_{2}$ by dividing the given mass by its molar mass.

$23 \cancel{\text{g CaC"_2xx(1"mol CaC"_2)/(64.0994cancel"g CaCl"_2)="0.358817711242 mol CaC"_2}}$ (I will round to two significant figures at the end. Meanwhile I'm keeping the calculator results until the end.

Convert $\text{mol CaC"_2}$$\rightarrow$$\text{mol H"_2"O}$ by multiplying the mol $\text{CaC"_2}$ by the mole ratio with water in the numerator.

$0.358817711242 \cancel{\text{mol CaC"_2xx(2"mol H"_2"O")/(1cancel"mol CaC"_2)="0.717635422484 mol H"_2"O}}$

Convert $\text{mol H"_2"O}$$\rightarrow$$\text{mass H"_2"O}$ by multiplying the mol $\text{H"_2"O}$ by its molar mass.

$0.717635422484 \cancel{\text{mol H"_2"O"xx(18.01528"g H"_2"O")/(1cancel"mol H"_2"O")=13"g H"_2"O}}$

We can put all of the steps together like this:

$23 \cancel{\text{g CaC"_2xx(1"mol CaC"_2)/(64.0994cancel"g CaC"_2)xx(2"mol H"_2"O")/(1cancel"mol CaC"_2)xx(18.01528"g H"_2"O")/(1cancel"mol H"_2"O")="13 g H"_2"O}}$

This way you can do the calculations all at once and round at the end, which reduces rounding errors.