Question

Suppose 29. g of butane is mixed with 144. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

butane = C4H10

the balanced equation is as follows

2 C4H10 + 13 O2 ------------------> 8 CO2 + 10 H2O

moles = mass / molar mass

for C4H10 => 29g / 58.12g/mol => 0.49896 moles

for O2 => 144 g / 32 g/mol =>4.5 moles

2mol C4H10 -----------------> 13 mol O2

0.49896 mol -------------------> ?

=> 0.49896 * 13 / 2 => 3.2432 moles of O2

limiting reactant is Butane

excess reactant is O2

so after the reaction the total butane is consumed and nothing is left over

answer => zero grams of butane ( or ) 0.0g of butane