How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made from 1.566 g of zn and excess hcl? the partial pressure of water vapor is 18.65 mm hg at 21.0°c.

a. 0.599 l?

The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.

1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn

0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced

Now use the ideal gas law to find the volume V.

P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm

T = 21 C + 273 = 294 K

PV = nRT

V = nRT / P = (0.02395 moles H2) (0.0821 L atm / K mole) (294 K) / (0.964 atm) = 0.600 L