20. mol H2 reacts with 8.0 mol O2 to produce H2O. Determine the number of grams reactant in excess and number of grams H2O produced. Identify the limiting reactant. 8.1 g H2 , 2.9 x 102 g H2O 17. How many litres of O2 gas are required to produce 100. g Al2O3?
How many moles of hydrogen gas can be produced when 10.0 g of Zn react with excess HCl? (Hint: you must write a balanced chemical reaction first.) 2. If the pressure is 1.2 atm and the temperature is 20.0ºC, what volume of hydrogen is produced in prelab question #1? 3.
What size metal duct should be used to deliver 270 CFM with a pressure drop of 0.15 in wc if the total equivalent length is 80 ft? (57.10) 8 in 5. What is the velocity of 500 CFM of air moving through a 10 in duct? (57.10) 900 fpm. 6.
B. How does the change in temperature affect the volume of the container? Get the Gizmo ready: Activity A: Boyle’s law Set the temperature (T) to 300 K. Check that the mass (m) is set to 0 kg. Question: How does pressure affect the volume of a gas? 1.
As the number of moles decreases, the volume decreases Summary: Combined Gas Law: PV/nT = constant (T in Kelvins) P1V1/n1T1 = P2V2/n2T2 Ideal gas Law: PV = nRT R = .0821L atm/mol K Gay-Lussac’s/Avogadro’s Law of Combining Volumes Equal volumes of any gases at the same temperature and pressure contain the same number of moles of gas. The coefficients of a balanced equation can be used to calculate relative volumes. Standard Molar Volume: At standard temperature and pressure (STP = 1atm and 273.15K) 1 mole of any ideal gas has a volume of 22.4L Variations on the ideal gas law equation: PV = mRT/M (m = sample mass, M = molar mass of the gas) d = MP/RT (d = density of the gas in g/L) Examples: 1. Calculate: a. The new pressure in a closed container if a 5.0L volume of gas at 2.5atm has its volume increased to 7.5L.
c. Prepare the solution by dissolving 38.90 grams of ZnI2 with 500 mL of water. d. 0.0125/0.25 = 0.05 L = 50 mL. This produces 0.0125 moles of ZnI2 5. Exercise 5: a. (0.125)(0.1) = 0.0125 moles of solute b. Pour 50 mL of the stock solution to get the number of moles needed.
Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
What is the final pressure in the drum if the final temperature is 60oC? A. 0.055 MPa B. 2.0 MPa C. 10 MPa D. 55 MPa Problems 35 and 36 are based on the following statement. Ten kilograms of oxygen and 2 kg of hydrogen are mixed in a 1 m3 vessel.
Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium. The reaction is as follows: Mg + 2HCl -> H2 + Mg2+ (aq) + 2Cl- (aq) There is a one to one relationship between the number of moles of hydrogen gas evolved and the
Empirical formula: CH5N Steps for molecular formula: 1- Calculate the molar mass of the empirical formula. 2- Divide the known (given) molar mass by the calculated empirical formula molar mass to get a whole number 3- Multiply that whole number through subscripts of the empirical formula to obtain the molecular formula. Example CH5N 12.01 g C x 1 C= 12.01 g/mol 1.008 g H x 5 H = 5.040