When a commercial drain cleaner that contains sodium hydroxide and small pieces

Question

When a commercial drain cleaner that contains sodium hydroxide and small pieces of aluminum is poured into a clogged drain, the following reaction occurs: Al + NaOH + H_2O rightarrow NaAl(OH)_4 + H_2 The heat generated by the reaction helps the sodium hydroxide break up the grease, and the hydrogen gas being generated stirs up the mixture and speeds up the unclogging of the drain. If 6.5 g Al and an excess of NaOH are used, what volume of gaseous H_2 measure at 742 mmHg and 22 degree C is produced?

Explanation / Answer

2 Al + 2 NaOH + 6 H2O ------------> 2 Na[Al(OH)4] + 3 H2

Mass of Al = 6.5 g.

Molar mass of Al = 27 g/mol

Number of moles of Al = mass / molar mass = 6.5 / 27 = 0.241 mol

from the balanced equation,

2 mol Al forms 3 mol H2

then, 0.241 mol Al forms 3 * 0.241 / 2 = 0.362 mol H2

SO, moles of H2, n = 0.362 mol

Temperature, T = 22 + 273 = 295 K

Pressure, P = 742 mmHg = 742 / 760 = 0.976 atm

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

Ideal gas equation,

P V = n R T

V = 0.362 * 0.0821 * 295 / 0.976

V = 8.98 L

V = 8980 mL

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