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Chemistry question? BEST ANSWER?
step by step please.
N2 + 2O2 yields 2NO2
Supposed 1.25 mol of N2 and 50.0g of O2 are mixed together.
A. What is the limiting reacant?
b. What is the maximum mass in grams of NO2 that can be produced from the mixture?
1 Answer
- Anonymous7 years agoFavorite Answer
m (O2) = 50 g
M (O2) = 32 g / mol
n (O2) = m / M = 50 / 32 = 1.56 mol
n (N2) = 1.25 mol
A)
Determining the limiting reactant:
According to the stoichiometry of the reaction:
1 mol of N2 reacts with 2 mol of O2
therefore, if we have 1.25 mol of N2 we will need "x" mol of O2 to completely react with the N2:
x = 1.25 x 2 = 2.5 mol of O2
since the actual amount of O2 available (1.56) is inferior than 2.5, then, O2 is the limiting reactant.
B)
Since O2 is the limiting reactant, we must work with it through the reaction:
Stoichiometric ratio (O2 / NO2): 2:2, so, the amount of NO2 formed is 1.56 mol.
M (NO2) = 46 g / mol
m (NO2) = 46 x 1.56 = 71.76 g