Question

Question #5f0a2

Answer 1

`"Grams of acetylene"` `~=` `2*g`

Explanation:

You have done the hard yards when you have the stoichiometric equation, which I note you have done:

`CaC-=C(s) + 2H_2O(l) rarr HC-=CH + Ca(OH)_2`

This equation explicitly says that `64.1*g` of calcium carbide react with `36*g` of water to produce `26*g` acetylene and `74*g` calcium hydroxide.

Each of these numbers represents molar equivalents of the reactants and products.

You started with `(5.00*g)/(64.1*g*mol^-1)` `=` `0.0780*mol` of calcium carbide. If follows that I formed an equivalent quantity of acetylene, i.e. `0.0780*mol`, and of calcium hydroxide.

So `"mass of acetylene = molar quantity"xx"molar mass"` `=` `0.0780*molxx26.0*g*mol^-1=??*g`

This reaction was the basis of the carbide lamp.