Question

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

`CaC_2 + 2H_2O -> Ca(OH)_2 + C_2H_2`

Answer 1

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

Explanation:

With water......

`Ca^(2+){C-=C}^(2-)+2H_2Orarr Ca(OH)_2+HC-=CH(g)uarr`

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

Is mass conserved in this reaction? Should it be?

And if you decompose `10* mol` calcium carbide you clearly gets `10* mol` acetylene. Agreed?

Answer 2

`"10 mols"` of `"C"_2"H"_2`.

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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 `"CaC"_2`, one mol of `"Ca"("OH")_2` is produced.

• For every one mol of `"CaC"_2`, one mol of `"C"_2"H"_2` is produced.

• For every one mol of `"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 `"CaC"_2` and `"C"_2"H"_2` have a `1:1` mol ratio. For emphasis, this is saying:

`color(red)(1)CaC_2 + 2H_2O -> Ca(OH)_2 + color(red)(1)C_2H_2`

So, you apparently already have the number written down. It is `color(blue)("10 mols")` of `"C"_2"H"_2` that is produced (assuming the water is in excess).