Moles of oxygen = mass / molecular weight = 90g / 32g/mol = 2.8125 mol From the

Question

Moles of oxygen = mass / molecular weight

= 90g / 32g/mol

= 2.8125 mol

From the stoichiometry of the reaction

3 mol O2 reacts with = 1 mol C2H4

2.8125 mol O2 reacts with = 2.8125/3= 0.9375 mol C2H4

O2 is limiting reactant

C2H4 is excess reactant

3 mol O2 produces = 2 mol CO2

2.8125 mol O2 produces = 2*2.8125/3 = 1.875 mol CO2

From the ideal gas equation

At STP

Pressure P = 1 atm

Temperature T = 273.15 K

Volume of CO2 = nRT/P

V = 1.875*0.0821*273.15/1

V = 42.05 L

Part b

C2H4 is excess reactant

Unreacted C2H4 = initial C2H4 - reacted C2H4

= 1.5 - 0.9375

= 0.5625 mol

Explanation / Answer

50. Gaseous ethene reacts with oxygen according to the equation: C2H4 (g) + 301g) 2co, (g) + 2H2 (g) a) What volume of carbon dioxide at STP can be produced when 1.50 moles of ethene reacts with 90.0g of oxygen? 3 n 1.bmol b) Which material is in excess? How much of it remains unreacted?

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