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.
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:
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.
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