Question

[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) ​

Answer 1

The mass of calcium carbide is 1,392 grams.

Step-by-step 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)

`n=(PV)/(RT)=(1.25 atm* 550 L)/(0.0821 atm L/mol K* 385 K)=21.75 mol`

`CaC_2(s) + 2 H2O (l)\rightarrow Ca(OH)_2 (s) + C_2H_2 (g)`

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:

`(1)/(1)* 21.75 mol=21.75 mol` 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.

Answer 2

The reaction is

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

As we have data of gas ethyne (or acetylene), C₂H₂

We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken

the moles of acetylene will be calculated using ideal gas equation

PV =nRT

R = gas constant = 0.0821 Latm/molK

T = 385 K

V = volume = 550 L

P = Pressure = 1.25 atm

n = moles = ?

n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol

As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide

moles of calcium carbide = 21.75mol

molar mass of CaC₂ = 40 + 24 = 64

mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g