Question

How many grams of O2 are neededto completely burn 41.8g of C3 H6?

Answer 1

2C₃H₆ + 9O₂ ------> 6CO₂ + 6H₂O

moles of C₃H₆ =
`(MASS)/(Molar Mass)`

=
`(41.8 g)/((12 * 3) + (6))`

= 0.995 mol

ratio of C₃H₆ : O₂
2 : 9

∴ mol of O₂ =
`(0.995 mol * 9)/(2)`

= 4.479 mol

Mass of O₂ = (Molar Mass) * (Mole)
= (16 * 2) g/mol * (4.479)mol
= 143.31 g