The equation is N2 + O2 reacts 2NO
Given 12.5 mg of oxygen reacts with 21.4 mg of Nitrogen what mass of product is formed
Do you have to take limiting factors into consideration yet?
In this case, your 12.5 mg O2 would produce 5.86 mg NO, and your 21.4 mg N2 would produce 11.5 mg NO. Therefore, O2 would be your limiting factor and this would produce 11.5 mg NO (or 0.0115 g NO)
If you need work, let me know
Probably you should show your work. I don't buy those numbers.
And yes, limiting reagent is a factor.
To determine the mass of the product formed, we first need to calculate the limiting reactant, which is the reactant that will be completely consumed in the reaction.
1. Calculate the molar masses of oxygen and nitrogen:
- Molar mass of O₂ = 2 * atomic mass of oxygen (16 g/mol) = 32 g/mol
- Molar mass of N₂ = 2 * atomic mass of nitrogen (14 g/mol) = 28 g/mol
2. Convert the given masses of oxygen and nitrogen to moles:
- Moles of O₂ = mass of oxygen / molar mass of O₂
= 12.5 mg / 32 g/mol
= 0.39125 mmol (millimoles)
- Moles of N₂ = mass of nitrogen / molar mass of N₂
= 21.4 mg / 28 g/mol
= 0.76428 mmol
3. Divide the number of moles of each reactant by its stoichiometric coefficient in the balanced equation to determine the ratio of moles used:
- Moles of O₂ used = moles of O₂ / coefficient of O₂ (which is 1)
= 0.39125 mmol / 1
= 0.39125 mmol
- Moles of N₂ used = moles of N₂ / coefficient of N₂ (which is 2)
= 0.76428 mmol / 2
= 0.38214 mmol
4. Compare the moles used for oxygen and nitrogen. The reactant with the smaller number of moles used is the limiting reactant.
In this case, oxygen (O₂) is the limiting reactant since it is completely consumed before nitrogen.
5. Determine the moles of the product (NO) formed:
- Since the stoichiometry of the balanced equation shows that for every 2 moles of nitrogen, 2 moles of nitrogen monoxide (NO) are formed:
- Moles of NO = moles of N₂ used (since the coefficient for NO is also 2)
= 0.38214 mmol
6. Convert the moles of NO to mass:
- Mass of NO = moles of NO * molar mass of NO
= 0.38214 mmol * (30 g/mol for nitrogen monoxide)
= 11.4642 mg
Therefore, the mass of product (NO) formed when 12.5 mg of oxygen reacts with 21.4 mg of nitrogen is approximately 11.4642 mg.