ovalentActivitydoll cato p-C CHEM 110 Spring 2016Sect. CengageBrain-My Home rain

Question

ovalentActivitydoll cato p-C CHEM 110 Spring 2016Sect. CengageBrain-My Home rain My HomeOWLv2|O .oma, Use the References to access impor For the following reaction, 24.4 grams of sulfur dioxide are allowed to react with 9.66 grams of oxygen gas. sulfur dioxide(g) + oxygen(g) sulfur trioxide(g) What is the maximum mass of sulfur trioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams Subinit Answer Retry Entire Group 3 more group attempts remaining

Explanation / Answer

For the first reaction:

SO2 + 1/2 O2 -> SO3

24.4 grams of SO2 * (1mol of SO2/64.066 g) = 0.3808 moles of SO2

9.66 grams of O2 * (1mol of O2/32 g) = 0.301875 moles of O2

Limiting reactant = SO2, as we need a half mole of oxygen per mole of sulfur dioxide

Maximum mass = 0.3808 moles of sulfur trioxide * (80 g/mol) = 30.464 grams

Mass of excess reagent = 0.1509375 moles * (32g/mol) = 4.83 grams of oxygen

For second reaction:

C + O2 -> CO2

7.85 grams of Carbon * (1mol / 12 grams) = 0.6542 moles of C

11.85 grams of Oxygen * (1mol / 32 grams) = 0.3703 moles of Oxygen

Limiting reactant = O2

Maximum mass of product = 0.3703 mol * (44 grams/mol) = 16.29375 grams

Mass of excess reagent = 0.2839 mol * (12 grams/mol) = 3.4068 grams

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