Question

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?

Answer 1

Approx. `13` `mL` of water.

Explanation:

Moles of calcium carbide: `(23*g)/(64.10*g*mol^-1)` `=` `0.359` `mol`.

Given the equation, `"2 equiv"` of water are required:

`0.359xx2*mol` water, `=` `0.718` `mol` water.

`"Equivalent mass"` `=` `0.718*molxx18.01*g*mol^-1` `=` `"approx. 12.9 g"` water.

Answer 2

You will need `"13 g H"_2"O"`.

Explanation:

Balanced Equation
`"CaC"_2("s")+"2H"_2"O(l)"` `rarr` `"C"_2"H"_2("g")+"Ca(OH)"_2("aq")"`

We will make the following conversions to answer this question.

`"mass CaC"_2` `rarr` `"mol CaC"_2` `rarr` `"mol H"_2"O"` `rarr` `"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

`"CaCl"_2":` `"64.0994 g/mol"`
https://pubchem.ncbi.nlm.nih.gov/compound/6352#section=Top

`"H"_2"O":` `"18.01528 g/mol"`
https://pubchem.ncbi.nlm.nih.gov/compound/962#section=Top

We also need the mole ratio between the two compounds.

From the balanced equation we can see that the mole ratio between `"CaC"_2"` and `"H"_2"O"` is `"1 mol CaC"_2: "2 mol H"_2"O"`.

Solution

Convert `"mass CaC"_2` `rarr` `"mol CaC"_2` by dividing the given mass by its molar mass.

`23cancel"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 `"mol CaC"_2"` `rarr` `"mol H"_2"O"` by multiplying the mol `"CaC"_2"` by the mole ratio with water in the numerator.

`0.358817711242cancel"mol CaC"_2xx(2"mol H"_2"O")/(1cancel"mol CaC"_2)="0.717635422484 mol H"_2"O"`

Convert `"mol H"_2"O"` `rarr` `"mass H"_2"O"` by multiplying the mol `"H"_2"O"` by its molar mass.

`0.717635422484cancel"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:

`23cancel"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.