A 2.50-g sample of impure sodium acetate was heated in the presence of excess so

Question

A 2.50-g sample of impure sodium acetate was heated in the presence of excess sodium hydroxide, producing methane gas and solid sodium carbonate. A volume of 415 mL of gas was collected over water at 27 degree C and 742 torr. a. What is the balanced chemical equation for the reaction? b. How many grams of methane were produced? The vapor pressure of water at 27 degree C is 27.8 torr. c. How many grams of sodium acetate are required to produce this quantity of methane? d. What is the percent pure sodium acetate in the impure sample?

Explanation / Answer

(a) Balanced equation is,

CH3COONa (aq.) + NaOH (aq.) -----------> CH4 (g) + Na2CO3 (s)

(b)

Volume of methane gas, V = 415 mL = 0.415 L

Pressure of methane gas, P = 742 - 27.8 = 714.2 torr = 714.2 / 760 = 0.940 atm

Temperature,T = 273.15 + 27 = 300.15 K

R = 0.0821 L.atm.K-1.mol-1

Molar mass of methan, M = 16 g/mol

Mass of methane, m = ?

Assuming ideal behaviour,

ideal gas equation, P V = n R T
P V = (m/M) R T

m = M P V / R T

m = 16 * 0.940 * 0.415 / (0.0821 * 300.15)

m = mass of methane produced = 0.253 g.

(c)

From the balanced equation,

1 mol (OR) 16 g. of methane is produced from 1 mol (or) 82 g. of sodium acetate

then, 0.253 g. of methane is produced from 0.253 * 82 / 16 = 1.30 g. of sodium acetate.

(d)

% sodium acetate = (1.30 / 2.50) * 100 = 52 %

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