There is either a typo or wrong post. You should have 36.26 g O2 remaining unreacted if all of the other numbers are right. Then grams on left = grams on right.
grams NH3 used + g O2 used = g N2 formed + g H2O formed
131.0................+ (221,2-36.26) = 107.8 + ..........g H2O
solve for g H2O. The answer is approx 200 g
You can do this another way.
mols NH3 = grams/molar mass = ?
Using the coefficients in the balanced equatiopn, convert mols NH3 to mols H2O.
Then convert mols H2O to gramss H2O by g H2O = mols H2O x molar mass H2O
Post your work if you run into trouble.
4NH3(g) +3O2(g) =2N2(g) +6H2O(l)
131.0 g of NH3 is mixed with 221.2 g of O2 and allowed to react. The reaction is complete when the NH3 is consumed. 107.8 g of N2 is formed and 36.60 g O2 remain unreacted. What mass of H2O is produced?
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