Question

Question 5 (Mandatory) (1 point) For the reaction shown, how many grams of oxygen will be required to react with 15.5 grams of chromium?4 Cr + 3 O2 --------> 2 Cr2O3Question 5 options:A) 12.7 g O2B) 21.5 g O2C) 14.3 g O2D) 7.15 g O2Question 6 (Mandatory) (1 point) Consider the following combustion reaction: 2 C4H10 + 13 O2 ------------> 8 CO2 + 10 H2OIf 5.05 grams of C4H10 reacts with more than enough of O2, what is the maximum amount of CO2 that can be formed?Question 6 options:A) 122 g CO2B) 15.3 g CO2C) 0.956 g CO2D) 61.2 g CO2Question 7 (Mandatory) (1 point) Consider the reaction between HCl and O2: 4 HCl + O2 ---------------> 2 H2O + 2 Cl2Calculate the theoretical yield of H2O when 64 grams of HCl react with 19 grams of O2.Question 7 options:A) 16 g H2OB) 21 g H2OC) 32 g H2OD) 43 g H2OQuestion 8 (Mandatory) (1 point) Consider the following chemical reaction: 4 Al + 3 O2 ---------------> 2 Al2O3If 12.85 g of Al reacts completely with oxygen and 18.42 g of aluminum oxide is obtained, what is the percent yield of the reaction?Question 8 options:A) 24.28%B) 75.86%C) 65.91%D) 69.76%Question 9 (Mandatory) (1 point) When a mole of gaseous propane (C3H8) reacts completely with oxygen gas, 2044 kJ of energy is formed, along with gaseous carbon dioxide and water vapor. Which of the following equations is a correct representation of this thermochemical reaction?Question 9 options:A) C3H8(g) + O2(g) -----------> CO2(g) + H2O(g) ΔH = -2044 kJB) C3H8(g) + O2(g) -----------> CO2(g) + H2O(g) + 2044 kJC) C3H8(g) + 5 O2(g) + 2044 kJ -----------> 3 CO2(g) + 4 H2O(g) D) C3H8(g) + 5 O2(g) -----------> 3 CO2(g) + 4 H2O(g) + 2044 kJQuestion 10 (Mandatory) (1 point) The evaporation of water is endothermic: H2O(l) + 44.01 kJ --------> H2O(g) If 275 kJ of heat is absorbed, what mass of water will evaporate?Question 10 options:A) 0.347 gB) 2.18 x 105 gC) 113 gD) 6.25 gQuestion 11 (Mandatory) (1 point) Based on the following reaction: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) If a reaction mixture contains 4.16 g of BaCl2 and 3.30 g of Na2SO4 how many moles of the precipitate will be formed?Question 11 options:A) 0.178 molesB) 0.0200 molesC) 0.0241 molesD) 0.0278 molesQuestion 12 (Mandatory) (1 point) The following chemical equation, which is unbalanced, shows the reaction for the rusting of iron: Fe(s) + O2(g) → Fe2O3(s)How many moles of O2 are required to react completely with 1.396 g of Fe?Question 12 options:A) 0.00937 moles of O2.B) 0.0187 moles of O2.C) 0.00469 moles of O2.D) 0.0375 moles of O2.Question 13 (Mandatory) (1 point) A reaction mixture of 18.215 g Na2CO3 and 10.938 g HCl react according to the following balanced equation: Na2CO3 + 2HCl → 2NaCl + H2O + CO2What is the theoretical yield for NaCl, in moles, from this reaction mixture?Question 13 options:A) 0.300 moles of NaClB) 0.172 moles of NaClC) 0.150 moles of NaClD) 0.344 moles of NaClQuestion 14 (Mandatory) (1 point) A reaction mixture of 0.1400 g of NaOH and 0.1400 g HCl reacts according to the following equation: NaOH + HCl → NaCl + H2O ∆H = -31.0 kJWhat is the heat change, q, corresponding to this reaction mixture?Question 14 options:A) 0.109 JB) -0.119 JC) -119 JD) -109 JQuestion 15 (Mandatory) (1 point) A mass of 0.6539 g of zinc is dissolved in excess hydrochloric acid, according to the following equation: Zn(aq) + 2HCl(aq) → ZnCl2(aq) + H2(g) How many moles of Cl- ions are formed in the reaction?Question 15 options:A) 0.002 molesB) 0.02 molesC) 0.01 molesD) 0.1 moles

Answer 1

Question 5

Answer

D - 7.15 g O2

Step-by-step explanation

Given:

Balanced chemical equation:

`4Cr\text{ + 3O}_2\rightarrow2Cr_2O_3`

mass of Cr = 15.5 g

Required: mass of Oxygen

Solution:

Step 1: Find the moles of Cr

n = m/M where n is the number of moles, m is the mass and M is the molar mass

n = 15.5 g/51,9961g/mol

n = 0.298 mol

Step 2: Use stoichiometry to find the moles of O

The molar ratio between Cr and O is 4:3

Therefore the number of moles of O = 0.298 x (3/4) = 0.224 mol

Step 3: From the moles, we can find the mass of O

m = n x M

m = 0.224 x 31.998

m = 7.168 g which is closer to D = 7.15 g