DAT Bootcamp- Stoichiometry and General Concepts – Flashcards

CrO - The trick to this problem is to recognize that the formula for percent weight of chromium (52 g/mol) in the sample is: This is a perfect example of where we plug in the answers given until something works. We don't have given masses, so we're going to use the atomic mass of chromium (52 g/mol) and the molar masses of the given answer choices until one of them equals 76.5%. Lets try CrO3 (52 + (3 x 16 = 48) = 100. So (52/100)*100 = 52% chromium, which isn't what we're looking for. Lets try CrO (52 + 16) = 68. (52/68)*100 = 76.5% chromium

1.5 mols - This is: 2NaN3 ->2 Na + 3N2 Recognize that nitrogen gas is diatomic! We see that sodium azide is 65g/mol from the periodic table, so that means only 1 mol of sodium azide was present. Thus, we can multiply the entire stoichiometric equation by 3/2, and find that 1.5 mols of N2 were produced 1 mol NaN3/ 1 x 3N2 / 2 NaN3 = 1.5 mols

151.5 - First, we need to write out the correct chemical reaction. This is: PbSO4 + Na2CrO4 -> PbCrO4 + Na2SO4. Thus, the molar ratio is 1:1. If we want to produce 0.162kg of lead chromate, then we need to find the number of mols produced. Lead chromate has a molar weight of 323 g/mol. Thus, 162g is approximately 0.50 mols. Thus, we need 0.5 mols of lead sulfate as well, which has a molar weight of 303g/mol. 303 x 0.50 = 151.5 PbSO4= 303g/mol PbCrO4 = 323g/mol 162g of Lead Chromate produced 162/323 = 0.50 mols 0.50 mols x 303g/mol = 151.5 g

(2) (160) / (380-185) - 2 g Liquid A / mL x 1/ (3801-185) g Liquid A x 160 g Liquid B / 1

0.25 mols, ammonium chloride - First, we calculate the number of mols of ammonium chloride, which has a molecular weight of 53. We find that there is 0.50 mol NH4Cl (26.5g / 53 g•mol-1). Sodium hydroxide has a molecular weight of 40g•mol-1, and thus 10g is 0.25 mols NaOH. We find that the stoichiometric ratio is 1:1 between NH4Cl and NH3, and 1:1 between NaOH and NH3. This means that NaOH is the limiting reagent, because all 0.25 mols of it will react with 0.25 mols of NH4Cl to produce 0.25 mols of NH3. This will leave an excess of 0.25 mols of ammonium chloride still unreacted (0.50 mols NH4Cl - 0.25 mols NH4Cl = 0.25 mols NH4Cl)

9 - The balanced equation for the combustion of ethanol (C2H6O) is: 1 C2H6O + 3 O2 → 2 CO2 + 3 H2O From the balanced equation, we find 1 + 3 + 2 + 3 = 9

C3O9 - The empirical formula is the lowest whole number ratio of the elements. The molar mass is the exact mass of the compound, which can be used to determine the compound's molecular formula. This can be done by dividing the molar mass by the empirical formula's mass, which gives you 3. This indicates that the subscripts in the empirical formula must be multiplied by 3 to obtain the molecular formula. CO3 = 60 g/mol 180 / 60 = 3x multiplier CO3 × 3 = C3O9

2, 13, 8, 10 -

3.21 x (180 / 339.8) - If the sample is 3.21 grams, you can determine the amount of water in the sample by calculating the percent composition of water and then multiplying that by the total grams in the sample. The percent composition of water is determined by adding the total grams of water in the compound = (18 g/mol x 10 mol) = 180 grams of water and dividing it by the sum of the atomic weights of all the elements in the compound, including water: CuSO4 • 10H2O = (63.54 + 32.06 + 4(16.00) + 180) = 339.8. %H2O (as a decimal) = mass of water/ mass of compound = 180/ 339.8

N7O2 - First to solve this problem, the empirical formula of the given compound must be determined. The mole ratio between the nitrogen and oxygen must be calculated by dividing by the lower mole number, 16 moles. 16/16 moles of oxygen = 1 mole oxygen 56/16 moles of nitrogen = 3.5 mole nitrogen This would yield the following empirical formula, N3.5O, however, the elements combine to form compounds using whole number ratios. We can't have 3.5 atoms of nitrogen. The empirical formula must therefore be multiplied by 2, resulting in N7O2.

(135 - 2P) / 3 - 3(atomic weight of X) + 2(atomic weight of P) = 135 g/mol. 3X + 2P = 135, so to solve for the atomic weight of unknown element X, the equation must be solved algebraically. Subtract 2P from both sides: 3X = 135 - 2P Then divide by 3X, giving you the atomic weight of element X = (135 - 2P) / 3

1.3 - The described reaction is the Haber process: N2 + 3 H2 → 2 NH3 This is a limiting reagent problem. We have to find out which reactant is going to run out first and limit our ability to produce ammonia gas. We can treat the liters of gas the same as mols in this case. This is because at a constant temperature and pressure, the volume of the gas is directly proportional to the number of mols. 2L H2 x 2 NH3 / 3 H2 = 4/3 = 1.33 mol NH3 2L N2 x 2 NH3/ 1 N2 = 4 mol NH3 So hydrogen gas is our limiting reagent in this case. Only 1.33 mols of ammonia can be produced before we run out of hydrogen gas and the reaction stops

5 - Hydrated salt problems can be tricky. Just remember this formula to find the value "x": mol H2O/ mol MgSO4 ="x" To find the mol of H2O, we subtract the original weight of the hydrated salt (the salt + water) by the weight of the anhydrous salt (only salt, no water). 21.0 - 12.0 = 9 grams of water. 9g / 18 (g•mol-1) = 0.5 mol of water *The molar mass of water is 18 g•mol-1. To find the mol of MgSO4, we use the anhydrous salt (only salt, no water). 12 / 120 = 0.10 mol of MgSO4 Finally, we plug back into the original equation to find: 0.5 / 0.1 = 5 So the formula of the hydrated salt is: MgSO4 • 5 H2O

silver, 2 cm3 - http://datbootcamp.com/wp-content/uploads/2013/09/Q3S3.png