For the following reaction, 0.390 moles of nitrogen gas are mixed with 0.139 moles of oxygen gas.
nitrogen (g) + oxygen (g) nitrogen monoxide (g)
What is the FORMULA for the limiting reagent?
What is the maximum amount of nitrogen monoxide that can be formed? moles
N2 + O2 ==> 2NO
How many moles NO would form if we used all of the N2 and all of the O2 we needed. That is 0.590 x (2 moles NO/1 mole N2) = 1.18 moles NO formed.
How many moles NO would form if we used all of the O2 and all of the N2 we needed. That will be
0.139 x (2 moles NO/1 mol O2) = 0.278 moles NO.
Both answer CAN'T be right, of course. The correct answer in limiting reagent problems is ALWAYS the smaller number. Therefore, O2 is the limiting reagent and 0.278 mol NO will be formed. If you want grams, that is g = moles x molar mass.
When a sample of potassium nitrate is heated, oxygen gas is produced. This gas, collected in a 750 milliliter flask, has a pressure of 2.8 atmosphere and the temperature is recorded to be 53.6°C. How many moles of oxygen gas are produced in this reaction?
(R = 0.08205 L atm/K mol)
YOUR ANSWER IS .078
To determine the limiting reagent, we need to compare the ratios of the moles of the reactants involved in the reaction to the stoichiometric coefficients in the balanced equation.
1. Write the balanced equation for the reaction:
nitrogen (g) + oxygen (g) → nitrogen monoxide (g)
2. Determine the stoichiometric coefficients for the reactants based on the balanced equation. In this case, the coefficient for nitrogen is 1, and the coefficient for oxygen is also 1.
3. Calculate the moles of each reactant involved:
moles of nitrogen = 0.390
moles of oxygen = 0.139
4. Calculate the ratio of moles of reactants to stoichiometric coefficients:
nitrogen: 0.390 / 1 = 0.390
oxygen: 0.139 / 1 = 0.139
5. Compare the ratios obtained. The limiting reagent is the one that has the smaller ratio. In this case, oxygen has the smaller ratio of 0.139 compared to 0.390 for nitrogen.
Therefore, the formula for the limiting reagent is oxygen (O2).
To calculate the maximum amount of nitrogen monoxide that can be formed, we need to use the stoichiometry from the balanced equation. The balanced equation shows that 1 mole of nitrogen reacts with 1 mole of oxygen to form 1 mole of nitrogen monoxide.
Since oxygen is the limiting reagent, we will use its moles to determine the maximum amount of nitrogen monoxide that can be formed.
moles of nitrogen monoxide = moles of oxygen = 0.139 moles
Therefore, the maximum amount of nitrogen monoxide that can be formed is 0.139 moles.