Urgent Chemistry Question

Question

If the previously shown reaction  begins with 76 g of 02 and 43 g of C2H6 identify the limiting reactant, the excess reactant, in the maximum number of grams of CO2 that can be formed. How many grams of excess reactant are left over? the chemical equation for this question is C2H6+ O2 —> H2O + CO2 which when balanced i believe is 2C2H6 + 7O2 —> 6H2O + 4CO2

Answer 1

2C₂H₆ + 7O₂ ==> 6H₂O + 4CO₂

moles O₂ present = 76 g x 1 mole/32 g = 2.375 moles O₂

moles C₂H₆ present = 43 g x 1 mole/30 g = 1.43 moles C₂H₆

Limiting reactant can be determined easily by dividing moles of substance by coefficient in the balanced eq. Thus...

O₂:  2.375/7 = 0.339

C₂H₆:  1.43/2 = 0.715

O₂ IS LIMITING

C₂H₆ is in EXCESS

 

Maximum CO₂ formed:

2.375 moles O₂ x 4 moles CO₂/7 moles O2 x 44 g/mole CO₂ = 59.7 g CO₂

 

C₂H₆ left over:

2 moles C₂H₆/7 moles O₂ x 2.375 moles O₂ = 0.679 moles C₂H₆ consumed or used

moles left = 1.43 - 0.679 = 0.75 moles left

grams left over = 0.75 moles x 30 g/mole = 22.5 grams left over