This is a limiting reagent problem. We know that because an amount is given for more than one of the reactants. This is a little more difficult than when only two reactants are present but I work these the same way.
Convert 1.24 kg propylene to mols. mols = grams/molar mass,
Convert 1.61 kg NH3 to mols.
Convert 1.90 kg O2 to mols.
Using the coefficients in the balanced equation convert mols propylene to mols acrylonitrile.
Do the same for NH3.
Do he same for O2.
It is likely that the values obtained for mols acrylonitrile will not agree which means two of them are wrong. The correct value in limiting reagent problems is ALWAYS the smallest value and the reagent producing that value is the limiting reagent (LR). Using that value, calculate grams acrylonitrile by g = mols x molar mass.
b.
Using the value of mols for limiting reagent, convert from mols LR to mols H2O. Then g H2O = mols H2O x molar mass H2O
2 C3H6(g) + 2 NH3(g) + 3 O2(g) → 2 C3H3N(g) + 6 H2O(g)
a. What mass of acrylonitrile can be produced from a mixture of 1.24 kg of propylene (C3H6), 1.61 kg of ammonia, and 1.90 kg of oxygen, assuming 100% yield?
b. What mass of water is produced?
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