How many grams of aluminum reacted in the following equation to produce 195 mL o

Question

How many grams of aluminum reacted in the following equation to produce 195 mL of hydrogen gas that was collected over water at a temperature of 19 degrees C and a pressure of 740 torr? The vapor pressure of water vapor at 19 degrees C is 16.0 torr 2NaOH+2Al+6H2O=2NaAl(OH)4+3H2 How many grams of aluminum reacted in the following equation to produce 195 mL of hydrogen gas that was collected over water at a temperature of 19 degrees C and a pressure of 740 torr? The vapor pressure of water vapor at 19 degrees C is 16.0 torr 2NaOH+2Al+6H2O=2NaAl(OH)4+3H2 2NaOH+2Al+6H2O=2NaAl(OH)4+3H2

Explanation / Answer

Solution:- given balanced equation is...

2NaOH + 2Al + 6H2O ---------> 2NaAl(OH)4 + 3H2

volume, V of H2 formed = 195ml = 0.195L

Temperature, T = 19 + 273 = 292K

Total pressure, Ptotal = 740 torr

water vapor pressure, Pwater = 16.0 torr

pressure of gas, Pgas = 740 torr - 16.0 torr = 724 torr

724 torr x 1 atm/760 torr = 0.953 atm

from ideal gas equation,

PV = nRT

n = PV/RT = 0.953 x 0.195/(0.0821 x 292) = 0.00775 moles

so, 0.00775 moles of H2 are formed.

Now, we will use the stoichiometry to calculate the grams of Al reacted.

0.00775 moles of H2 x (2 mol Al/ 3 mol H2) x (26.98g Al/1mol Al) = 0.139g of Al.

So, the answer is 0.139 g Al.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.