If a lamp uses 975 mL of C2H2 per hour at 1.00 atm with a temperature of 291.15 K, how many grams of CaC2 must be in the lamp for an 8 hour shift?
975 mL/hr x 8 hr = 7800mL = 7.8 L
Using PV = nRT
(1 atm) (7.8 L) = n (0.8206) (291.15 K)
n = 0.326 moles C2H2
multiply by molar mass
0.326 moles x 26 g/mole = 8.48 g
Comments
975 mL/hr x 8 hr = 7800mL = 7.8 L
Using PV = nRT
(1 atm) (7.8 L) = n (0.8206) (291.15 K)
n = 0.326 moles C2H2
multiply by molar mass
0.326 moles x 26 g/mole = 8.48 g