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?
Answer: 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
