Question

For the following reaction, 10.5 grams of butane (C4H10) are allowed to react with 20.4 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? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining

Answer 1

Answer

Maximum amount of carbondioxide that can be formed : 17.27g

what is the formula for limiting reagent : O2

what amount of excess reagent remains after reaction is complete : 4.798g

Explanation

The balanced equation is

2C4H10(g) + 13O2(g) ------> 8CO2(g) + 10H2O(g)

stoichiometrically, 2moles of C4H10 reacts with 13 moles of O2

mole = mass/molar mass

given moles of C4H10 = 10.5g/58.14g/mol = 0.1806mol

given moles of O2 = 20.4g/ 32g/mol = 0.6375mol

0.1806moles of C4H10 need 1.1739moles of O2 but available moles of O2 is 0.6375mol

Therefore,

Limiting reagent is O2

stoichiometrically, 13 moles of O2 gives 8moles of CO2

number of moles of CO2 obtained by 0.6375moles of O2 = (8/13) × 0.6375mol = 0.3923mol

maximum amount of CO2 that can be formed = 0.3923mol × 44.01g/mol = 17.27g

moles of C4H10 react with 0.6375moles of O2 = ( 2/13)× 0.6375mol = 0.09808mol

Moles of C4H10 after completion of reaction = 0.1806mol - 0.09808mol = 0.08252mol

mass of C4H10 remaining after completion of reaction = 0.08252mol × 58.14g/mol = 4.798g