Question

a. A sample of 1.41 g of helium and an unweighed quantity of O2 are mixed in a flask at room temperature. The partial pressure of helium in the flask is 42.0 torr , and the partial pressure of oxygen is 159 torr. What is the mass of the oxygen in the container?

b.  Both Jacques Charles and Joseph Louis Guy-Lussac were avid balloonists. In his original flight in 1783, Jacques Charles used a balloon that contained approximately 31100 L of H2. He generated the H2 using the reaction between iron and hydrochloric acid: Fe(s)+2HCl(aq)→FeCl2(aq)+H2(g). How many kilograms of iron were needed to produce this volume of H2 if the pressure was 1.00 atm and the temperature was 25 ∘C?

Answer 1

Given that 1.41 g of helium and an unweighed quantity of O2 partial pressure of helium =42.0 torr partial pressure of oxygen = 159 torr total pressure = 42.0+159=201 torr mole fractions 1.41gHe/4g/mole =0.353moles He mole fraction He x total pressure = partial pressure He ?torr x 201 torr = 42.0 torr 0.208 = mole fraction of He total moles of gas present: 0.353 moles He /total moles = 0.208 total moles = 1.69 moles 1.69 moles total - 0.353 moles He=moles 02 = 1.337 moles 02 1.337moles x 32g/mole = 42.7 g 02 Fe(s)+2HCl(aq) ---- > FeC12(aq)+H2(g). Now calculate the number of moles of gas by ideal gas law as follows: PV=nRT P= pressure, 1.00 atm T= 25 C or 298 K V=31100 L Number of moles is calculated as follows: n =PV/RT = 1.00*31100/0.0821*298 = 1271.16 moles H2 Now calculate the mole of Fe: 1271.16 moles H2 *1 mole Fe/ 1 mole H2 =1271.16 moles Fe Amount in g = 1271.16 moles Fe * molar mass = 1271.16 moles Fe * 55.845 g/mole = 70987.9302 g Fe = 70.99 kg Fe