D Acetylene C2H2 can be prepared by combining calcium carbid

D. Acetylene, C2H2, can be prepared by combining calcium carbide, CaC2, and water according to the following reaction. CaC2 (s) 2H20 (e) C2H2 (g) Ca(OH)2 (s) To produce the acetylene, the procedure suggests that 50.0% more water than re- quired be added to the calcium carblde. What mass of calcium carbide and volume of water (d 1.00 gmLI must be combined according to the procedure, so that 10.0 L of acetylene at 27°C and 1.00 atm are produced?

Solution

Balanced equation:
CaC₂ + 2 H₂O = C₂H₂ + Ca(OH)₂

Using ideal gas equation we have find moles acetylene produced

PV = nRT

P = 1 atm , V = 10 L, n =?

R = 0.0821 L atm K-1 Mole-1 T = 300 K

n = 1 x 10 / 0.0821 x 300 = 0.406 Moles

0.406 Moles of acetylene was produced

To produce 0.406 Moles of acetylene we need the same moles of CaC2 and 1.218 Moles of Water (Including extra 50%)

0.406 Moles of CaC2 = 0.406 x 64.09 = 26.02 gm

1.218 Moles of Water = 1.218 x 18 = 21.92 gm

Hence 26.02 gm of CaC2 and 21.92 ml of water is need to produce 10 l of acetylene at 27C

Question 4

First let us find energy needs to to reach 0⁰C of all the soft drink

q = m x C x (T_f - T_i)

q = amount of heat energy gained or lost by substance

m = mass of sample

C = heat capacity (J ^oC^-1 g^-1 or J K^-1 g^-1)
T_f = final temperature
T_i = initial temperature

q = 361 x 4.18 x 23 = 34706 Joules are requird

Now let us calculate the weight of ice = 34706 /334 = 103.91

Hence 103.91 gm of ice had melt