250. grams of ethane are allowed to react with 220. grams of oxygen gas. Which of these
reactants are the limiting reactant, and how many grams of carbon dioxide would be produced?

1) Ethane is the limiting reactant

2) mass of CO₂ produced is 733.33 g

Explanation:

C₂H₆ + 2O₂ → 2CO₂ + 3H₂ (1)

molar ratio for equation (1) can be shown as;

1 : 2 → 2 : 3

molecular weight of moles of C₂H₆ = 30 g/mol

molecular weight of moles of O₂ = 32 g/mol

molecular weight of moles of CO₂ = 44 g/mol

mass of O₂ = 220 g

mass of C₂H₆ = 250 g

number of moles of O₂ = mass ÷ molecular weight = 220 g ÷ 32 g/mol * 2 = 13.75 moles

number of moles of C₂H₆ = mass ÷ molecular weight = 250 g ÷ 30 g/mol = 8.333 moles

1) Ethane is the limiting reactant as the 8.333 moles will finish before the 13.75 moles of the oxygen is totally consumed.

2) 1 mole of C₂H₆ will form 2 moles of CO₂

Therefore, 8.333 moles will form 16.667 moles of CO₂

mass of CO₂ formed = number of moles * molecular weight = 16.667 * 44 = 733.33 g

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

lol for the pointsss