For the following reaction, 4.45 grams of butane (C4H10) are allowed to react with 19.0 grams of oxygen gas . butane (C4H10)(g) + oxygen(g) carbon dioxide(g) + water(g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? O2 What amount of the excess reagent remains after the reaction is complete? grams

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thomasdecabooter thomasdecabooter · Answer 1 · 2020-03-25T04:08:49+01:00

Answer:

13.5 grams of CO2 can be produced. C4H10 is the limiting reactant. O2 is in excess. There remains 3.08 grams

Explanation:

Step 1: Data given

Mass of butane = 4.45 grams

Molar mass of butane = 58.12 g/mol

Mass of O2 = 19.0 grams Molar mass of O2 = 32.0 g/mol

Step 2: The balanced equation

2C4H10 + 13O2 → 8CO2 + 10H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles butane = 4.45 grams / 58.12 g/mol

Moles butane = 0.0766 moles

Moles O2 = 19.0 grams / 32.0 g/mol

Moles O2 = 0.594 moles

Step 4: Calculate the limiting reactant

For 2 moles butane we need 13 moles O2 to produce 8 moles CO2 and 10 moles H2O

Butane is the limiting reactant. It will completely be consumed (0.0766 moles). O2 is in excess. There will react 6.5 * 0.0766 = 0.4979 moles

There will remain 0.594 - 0.4979 = 0.0961 moles

This is 0.0961 * 32.0 = 3.08 grams

Step 5: Calculate moles CO2

For 2 moles butane we need 13 moles O2 to produce 8 moles CO2 and 10 moles H2O

For 0.0766 moles butane we'll have 4*0.0766 = 0.3064 moles CO2

Step 6: Calculate mass CO2

Mass CO2 = 0.3064 moles * 44.01 g/mol

Mass CO2 = 13.5 grams

13.5 grams of CO2 can be produced. C4H10 is the limiting reactant. O2 is in excess. There remains 3.08 grams