based on the following chemical equation
HCN+O2 yields N2+CO2+H20
identify the limiting reactants and the mass of N2 produced when 100.0g of HCN react with 100.0g of O2.
a. the limiting reactant is HCN and 25.9g of N2 are produced
b. the limiting reactant is O2 and 35.0g of N2 are produced
c. the limiting reactant is HCN and 51.8g of N2 are produced
d. both reactants are consumed completely
Here is a worked example of how to solve limiting reagent problems. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
Got it....thanks Dr Bob
To determine the limiting reactant and the mass of N2 produced when HCN and O2 react, we need to perform a stoichiometric calculation.
First, we need to convert the given masses of HCN and O2 to moles using their respective molar masses. The molar mass of HCN is 27.03 g/mol, and the molar mass of O2 is 32.00 g/mol.
Number of moles of HCN = mass of HCN / molar mass of HCN
= 100.0 g / 27.03 g/mol
≈ 3.7 mol
Number of moles of O2 = mass of O2 / molar mass of O2
= 100.0 g / 32.00 g/mol
≈ 3.1 mol
Next, we need to use the stoichiometry of the balanced equation to determine which reactant is the limiting reactant. The coefficients in the balanced equation give us the mole ratio between the reactants.
The balanced equation states that 1 mole of HCN reacts with 1 mole of O2 to produce 1 mole of N2. Therefore, the mole ratio of HCN to O2 is 1:1. Based on our calculations, we can see that there is an equal molar amount of HCN and O2, suggesting that either reactant could be the limiting reactant.
To decide which reactant is the limiting reactant, we need to find out how many moles of N2 can be produced from each reactant. We can assume that each reactant will react completely according to the stoichiometry of the balanced equation.
Using the mole ratio from the balanced equation, we can calculate the moles of N2 produced from the given amount of HCN:
Number of moles of N2 produced (from HCN) = moles of HCN
= 3.7 mol
Similarly, we can calculate the moles of N2 produced from the given amount of O2:
Number of moles of N2 produced (from O2) = moles of O2
= 3.1 mol
Now, we compare the moles of N2 produced from each reactant. The reactant that produces the least number of moles of N2 is the limiting reactant.
In this case, 3.1 mol is less than 3.7 mol. Hence, O2 is the limiting reactant.
Finally, calculate the mass of N2 produced from the limiting reactant. We can use the mole to mass conversion using the molar mass of N2, which is 28.01 g/mol.
Mass of N2 produced = moles of N2 (from O2) * molar mass of N2
= 3.1 mol * 28.01 g/mol
≈ 86.5 g
Therefore, the correct answer is:
b. The limiting reactant is O2 and 35.0g of N2 are produced.