Questions 6-7 refer to manganese (VII) oxide (Mn_2O_7, MM = 221.9 g/mol) 82-3 mi
ID: 531668 • Letter: Q
Question
Questions 6-7 refer to manganese (VII) oxide (Mn_2O_7, MM = 221.9 g/mol) 82-3 milligrams of Mn_2O_7, corresponds to how many millimoles of Mn_2O_7? A) 0.371 millimoles B) 371.0 millimoles C) 2.696 millimoles D) 269.6 millimoles How many moles of oxygen atoms (0) are there in 2.45 moles of manganese (VII) oxide? A) 2.45 moles B) 4.90 moles C) 17.15 moles D) 34.3 moles Questions 8-10 refer to the production of acrylonitrile (C_3H_3N, MM = 53 g/mol) from propylene (C_3H_6, MM = 42 g/mol) according to the (balanced) reaction: 2C_3H_6 + 2NH_3 + 30_2 rightarrow 2C_3H_3N + 6 H_2O The chemical industry typically uses approximately 20 times 10^9 kg of propylene (C_3H_6) to produce acrylonitrile (C_3H_3N) each year. What is the maximum amount of acrylonitrile that can be produced from this amount of starting material? A) 2.52 times 10^9 grams B) 2.52 times 10^10 grams C) 2.52 times 10^12 grams D) 2.52 times 10^13 grams How many moles of oxygen are needed to produce the acrylonitrile in # 8? A) 7.14 times 10^11 B) 7.14 times 10^10 C) 7.14 times 10^11 D) 7.14 times 10^7 This reaction actually has many byproducts and thus does not occur with a 100% yield. If the ordering department of the factory that produces acrylonitrile ordered 4.5 times 10^9 kg of propylene (1.07 times 10^11 moles) and 3.2 times 10^9 kg of ammonia (1.88 times 10^11 moles) and then used those reactants (in excess oxygen) to produce 3.8 times 10^9 kg of acrylonitrile, what was the percent yield of the reaction? A) 38% B) 67% C) 149% D) 355%Explanation / Answer
6.Number of millimoles= mass in mg / molar mass
= 82.3 mg/ 222 (g/mol)
=0.371 mmol
Therefore option (A) is correct.
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