[15 points, stoichiometry and gases]

find the grams of cac2 (@ 1.25 atm, 385 k and c2h2 has 550 l)

cac2 (s) + 2 h2o (l) > ca(oh)2 (s) + c2h2 (g)

The mass of calcium carbide is 1,392 grams.

Explanation:

Pressure of the ethyne gas = P = 1.25 atm

Volume of the ethyne gas = V = 550 L

Temperature of the ethyne gas = T = 385 K

Moles of ethyne gas = n

PV = nRT (ideal gas equation)

According to reaction , 1 mole of ethyne is obtained from 1 mole of calcium carbide.

Then 21.75 moles of ethyne will be obtained from:

calcium carbide.

Mass of 21.75 moles of calcium carbide:

= 21.75 mol × 64 g/mol = 1,392 g

The mass of calcium carbide is 1,392 grams.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3