Underground miners formerly used headlamps in which calcium carbide (CaC2) was reacted with water to form acetylene (C2H2) gas as shown in the following reaction:

Question

CaC2(s) + H2O (l) ---> C2H2 (g) +Ca(OH2)(s)

The acetylene produced was then burned to produce a very bright light. Starting with 84.0g of calcium carbide and excess water, what volume of acetylene will be produced at 300 K and 1520 torr?

Now I balanced the reaction so that it becomes this:
CaC2(s)+2H2O(l)-->C2H2(g)+Ca(OH2)(s)

and 1520 torr equals about 2 atm but I'm not really sure where else to go from here.

Answer 1

I have corrected your equation.
CaC2(s)+2H2O(l)-->C2H2(g)+Ca(OH)2(s)

mols CaC2 = 84g/molar mass = ?
Using the coefficients in the balanced equation, convert mols CaC2 to mols C2H2.

Then convert mols C2H2 to volume at the conditions listed using PV = nRT and solve for V in liters. For P you will need to convert 1520 torr to atm.