Name Stoichiometry EC Due by Friday Nov 17 in class Worth 20 quiz points 1. Acry

Question

Name Stoichiometry EC Due by Friday Nov 17 in class Worth 20 quiz points 1. Acrylonitrile (C,H;N) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction: 2C3H6 (g) + 2 NI 13 (g) + 3 O2 (g) 2CJI,N (g) + 6 H2O (g) 15.0g C,Ho, 10.0g O2 and 5.00g NHj are reacted. A. Determine the limiting reagent. B. Calculate the theoretical yield of C,H,N in grams. C. Determine the mass in grams of both of the excess reagents that will be left over. D. What is the percent yield if 10 g of acrylonitrile is obtained?

Explanation / Answer

1.

We have reaction equation as;

2 C3H6(g) + 2 NH3(g) + 3 O2(g) = 2 C3H3N(l) + 6 H2O(g)

mass of propene= 15g

mass of oxygen = 10g

mass of ammonia = 5g

calculate moles of each by dividing with their respective molar masses

propene moles = 15g/ 42g/mol = 0.3571 moles

oxygen moles = 10g/32g/mol = 0.3125 moles

ammonia moles = 5g/17g/mol = 0.2941 moles

now use stoichiometric conversion factor to find the amount of product formed:

acrylonitrile from propene = 2 mole acrylonitrile / 2mol propene x 0.3571 moles propene = 0.3571 moles acrylonitrile

acrylonitrile from oxygen = 2 mole acrylonitrile / 3mol oxygen x 0.3125 mol oxygen = 0.2083 moles acrylonitrile

acrylonitrile from ammonia = 2 mol acrylonitrile / 2 mol ammonia x 0.2941 mol ammonia = 0.2941 mol acrylonitrile

The one which gives least product is limiting reactant and the reaction will stop there as the limiting reactant is over.

A. oxygen is limiting reactant

B. theoretical yield will be based on limiting reactant

we have 0.2083 mol acrylonitrile

mass = moles x molar mass = 53 g/mol x 0.2083 mol = 11.04 g

C.ammonia and propene are excess reactants

stoichiometric conversion factor for propene to oxygen = 2 mol propene / 3 mol oxygen

moles of propene used = 2 mol propene / 3 mol oxygen x 0.2083 mol oxygen = 0.1389 moles propene

propene left = 0.3571 mol - 0.1389 mol = 0.2182 mol

mass left = 0.2182 mol x 42 g/mol = 9.1658 g

similarly for ammonia = 2 mol ammonia / 3 mol oxygen x 0.2083 mol oxygen = 0.1389 moles ammonia

moles of ammonia left = 0.2941 mol - 0.1389 mol = 0.1552 mol ammonia

mass = 0.1552 mol x 17g/mol = 2.6384 g

D. % yield = actual yield / theoretical yield x 100 = 10g/11.04 g x 100 = 90.5797 ~ 90.6 %

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