Question

Some commercial drain cleaners contain two components: sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: 2NaOH(aq) + 2Al(s) + 6H2O(l) 2NaAl(OH)4 (aq) + 3H2(g) The heat generated in this reaction helps to melt away grease and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of H2 formed at 298 K and 1.1 atm if 1.34 g of Al is treated with excess NaOH and H2O

Answer #1

The balanced equation is-

2NaOH (aq) + 2Al(s) + 6H2O(l) = 2NaAl(OH)4 (aq) + 3H2(g)

So, 3 moles hydrogen is produced from 2 moles of Aluminium.

Molar Mass of Al = 27 g/mol. Therefore, 1.34 g Al = 1.34/27 mol

Moles of H2 gas produced = (3/2)*(1.34/27) = 0.0744

From the combined gas laws we know that,

PV = nRT

Or, V = nRT/P, where, P = 1.1 atm, T= 298 K, R = 0.082 L atm / mol K, n = 0.0744 mol

By substituting the values, we get,

V = 0.0744*0.082*298/1.1 L = 1.653 L

Volume of H_{2} formed = 1.653 L

Aluminum reacts with aqueous sodium hydroxide to produce
hydrogen gas according to the following equation: 2Al(s) +
2NaOH(aq) + 6H2O(l)2NaAl(OH)4(aq) + 3H2(g) The product gas, H2, is
collected over water at a temperature of 20 °C and a pressure of
750 mm Hg. If the wet H2 gas formed occupies a volume of 9.03 L,
the number of grams of H2 formed is g. The vapor pressure of water
is 17.5 mm Hg at 20 °C.

Aluminum reacts with aqueous sodium hydroxide to produce
hydrogen gas according to the following equation: 2Al(s) +
2NaOH(aq) + 6H2O(ℓ) 2NaAl(OH)4(aq) + 3H2(g) In an experiment, the
H2 gas is collected over water in a vessel where the total pressure
is 758 torr and the temperature is 20°C, at which temperature the
vapor pressure of water is 17.5 torr. Under these conditions, the
partial pressure of H2 is __torr. If the wet H2 gas formed occupies
a volume of 9.60...

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