Question

Lab 6: Chemical Reactions and Equations AND Lab 9: Aspirin (stoichiometry for both): 1. Given the following equation: 2 C4H10 + 13 02 ---> 8 CO2 + 10 H20, show what the following molar ratios should be. Also- what type of reaction is this? a. C4H10/02 b. 02/CO2 c. 02/H2O d. C4H10/CO2 e. C4H10/H2O 2. Given the following equation: 2 KClO3 ---> 2 KCl + 3 Oz How many moles of O2 can be produced by letting 12.00 moles of KCIO3 react? How many O2 molecules is that? What type of reaction is this? 3. Given the following UNBALANCED equation: K+ Cl ---> KCI a. How many grams of KCl is produced from 2.50 g of K and excess Clz. b. From 1.00 g of Cl, and excess K? c. Also- what type of reaction is this? Is it also a redox reaction? If so, which starting material is reduced?

Answer 1

1. Aus - The given reaction is 204 Ho + 1302 80 +1000. I @ 2 modes of cattio reacts with in moles of a I 30 the molar ratio of Caltio 102 is 2:13. 6 13 mous of Oz produces 8 moles of Co2 so, the molar ratio of Oal coa is 18/8.. 02 m)Сано редкая 8 mm) 4 Ф, so, the molar ratio of cattol co2 in 2/8 . - = - = 10 4

1 @ 2 moes of sto produces 10 mois of 10 So, the molar ratio is 2:10 = 1:5 This is a is combusted combustion to producid reaction. In this reaction co2 and 10. Calto 2. Ans - The given reaction is 2 kun -> QKU + 302 This is a balanced equation, from the balanced equation, а ku0, реда, 2 ms , 12 moles of klos will produce 3x12 modes of Oz - 18 miles of os I mole of of contains 6.022 x 1823 molecules :, 18 miles of Oz contains 6.022 x 1823 x 18 molecules of O2 = 108.396x10 23 02 molecules. This is a decomposition reaction Kuo, decomposes to form kd and o

and produces 13. Ans. The given reaction is K+Ua ka The balanced equation is - 2k + Cl2 → akd. from the balanced equation, 2 mous of k reacts with Imole of u 2 mous of kl. 2 moles of k contains 2x39.09 g of k = 78-18 g of k 2 moles of ku contains 2x74.55 g of ko = 149.1 g of ka. @ Ans - as I is present in exces, so the product will depend on the amount of k. 78.18 g of k produces 14901 g of ka na oft will produces 149.1 x 2.50 g of ka 78.18 = 4.767 g of ka. 7 product will Ans - b k dipind on I mole of is the a present in excen, so the amount of 2. contains 2x35.45 g of Uz = 70.9 9 of ₂

Bon the balanced equation. 70.9 g of u produces 149.1g of ka l g of U will produce 19901 g of ku T - 2.109 of ku. @ Ars - Yes it is a redox reaction. This is a combination reaction from the balanced equation, the oxidation no. of k from 0 to +1 and a change from 0 to 1 . So U change is reduced.