I need help with this question please!! And quickly if at all possible... H2 gas

Question

I need help with this question please!! And quickly if at all possible...

H2 gas is produced in the reaction between Al metal and commercial drain cleaner containing NaOH. The heat generated helps NAOH to break up rhe grease. If 6.5 g of Al and an excess of NaOH are used, what volume of gaseous H2 is measured at 742 mm Hg and 22 degrees Celsius is produced? Remember H2 (g) is collected over water at 22 degrees Celsius. Al (s) + NaOH (aq) + H2O (l) ----> NaAl(OH)4 (aq) + H2 (g) (At 22 degrees Celsius the vapor pressure of the water is 26.7 torr.)

Explanation / Answer

1 mole of Al = 27 g of Al gives 1 mole of H2 = 2 g

Thus, 6.5 g will give: 6.5 * 2 / 27

= 0.48148 g of H2 or 0.24074 moles of H2

Now, using the ideal gas equation,

V = nRT / P

P = 742 mm Hg = 742 / 760 = 0.97631 atm

n = 0.24074

R = 0.0821

T = 22 + 273 = 295 K

Thus, V = 0.24074 * 0.0821 * 295 / 0.97631

= 5.972 L of H2 is collected.

Hope this helps. Please rate it ASAP. thanks.

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