I need to see the work to figure out where I am going wrong... Acetelene, C2H2 ,

ID: 633261 • Letter: I

Question

I need to see the work to figure out where I am going wrong...

Acetelene, C2H2, can be prepared by combining calcium carbide, CaC2, and water according to the following reaction.

CaC2 (s) + 2 H2O (l) -----------> C2H2 (g) + Ca(OH)2 (s)

To produce the acetylene, the procedure suggests that 50.0% more water than required be added to the calcium carbide.

* What mass of calcium carbide and volume of water (D = 1.00 g/mL) must be combined according to the procedure,

so that 10.0 L of acetylene at 27 degrees C and 1.00 ATM are produced?

For acetylene, PV = nRT

1.00 x 10.0 = n x 0.082 x 300

n = moles of acetylene = 0.407 moles.

1 mole of acetylene is given by 1 mole of calcium carbide. Thus moles of carbide = 0.407

mass of carbide = moles x molecular mass = 0.407 x 64 = 26.048 gms.

mass of water required = 2 x 0.407 x 18 gms = 14.652 gms. Actual mass of water = 14.652 + 0.50 x 14.652 = 21.978 gms.

Thus, volume = 21.978 ml as D = 1.00 g/mL