CHM Exam 5
A 0.630 gram sample of a metal, M, reacts completely with sulfuric acid according to: M(s)+H2SO4(aq)-->MSO4(aq)+H2(g) A volume of 291 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 Torr and the temperature is 25 °C. Calculate the molar mass of the metal.
1) find pressure H2 P(h2)=Ptotal - Pwater =756 torr-23.8 torr =732.2 torr 2) find moles H2 PV=nRT (732.2 torr)(291 mL x 1 L/1000 mL)=n(0.08206)(298 K) 213.07 torr x L = 24.45 n = 8.714 torr x (1 atm/760 torr) moles H2 =0.011466 moles M(s)=.011466 x (1 mol M(s)/1 mol H2)=.011466 mol M(s) molar mass M(s)=.630 g/4.32 mol =0.146
How much heat energy is required to convert 51.5 g of solid iron at 28 °C to liquid iron at 1538 °C? The molar heat of fusion of iron is 13.8 kJ/mol. Iron has a normal melting point of 1538 °C. The specific heat capacity of solid iron is 0.449 J/g·°C.
q1=(m)(s)(deltaT) =(51.5 g)(.000449 kJ)(1538 C-28 C) =34.92 kJ q2=(n)(heat of fusion) # moles Fe(s)=51.5 g/ 55.845 mol mass =0.922 mol Fe(s) q2=(0.922 mol)(13.8 kJ) =12.7 kJ q=q1+q2 =34.94+12.7 --> 47.64 kJ <--
The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 2.87 g of water boils at atmospheric pressure?
q= (n)(heat of vaporization) moles H2O= 2.87 g/40.66g =0.159 moles q=(0.159 mol)(40.66 kJ/mol) =6.48
phase change formula
q=(amount)(enthalpy change)
formula for calculating the heat necessary to raise the temperature of M(s) to its melting point
q=(m)(s)(deltaT) m=mass given in reaction s=specific heat of given substance deltaT=temperature change (total temp-temp of substance)
formula for finding amount of heat necessary
q=(n)(heat of fusion or heat of vaporization)
At 1 atm, how much energy is required to heat 69.0 g of H2O(s) at -10.0 °C to H2O(g) at 121.0 °C?
the solid water (ice) must go through a liquid and a gas change.You must account for the individual temp changes (q=M x C x deltaT) and the phase changes (q=amount x deltaH) separately. DeltaH is the enthalpy change and C is the specific heat (of either ice, water, or steam). 1) temp change ice=(69 g)(2.087 J)(10 C) =1440.03 J -----> (10 Celsius was used because you must calculate degrees to get to zero Celsius. Since you started at -10 Celsius it takes +10 to get to 0 Celsius) 2) phase change ice=(69 g)(333.6 J) =23018.4 J 3) temp change water=(69 g)(4.184 J)(100 C) =28869.6 J --------> (100 Celsius was used because water turns to steam at this temperature) 4)phase change water=(69 g)(2257 J) =155733 J 5) temp change steam=(69 g)(2.00 J)(21 C) =2898 J q=q1+q2+q3+q4+q5 then convert Joules to Kilo-joules --> 211.96 kJ <--
