18.0 grams of oxygen is reacted with nitrogen to make 56.0 grams of nitrogen dioxide according to the following reaction:
2O2 + N2 ---> 2 NO2
How much nitrogen (in grams) reacted?
To find out how much nitrogen reacted, we need to determine the number of moles of nitrogen that reacted in the reaction. We can do this by using the molar masses and the stoichiometry of the reaction.
First, we need to determine the number of moles of oxygen reacted. We can use the given mass of oxygen (18.0 grams) and its molar mass to calculate this:
Molar mass of oxygen (O2) = 16.00 g/mol
Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
= 18.0 g / 32.00 g/mol
= 0.5625 mol
Next, we can use the stoichiometry of the reaction to determine the number of moles of nitrogen reacted. From the balanced equation, we can see that the stoichiometric ratio of oxygen to nitrogen is 2:1:
2O2 + N2 ---> 2NO2
So, for every 2 moles of oxygen reacted, 1 mole of nitrogen reacts.
Number of moles of nitrogen = (2/1) * Number of moles of oxygen
= (2/1) * 0.5625 mol
= 1.125 mol
Finally, we can calculate the mass of nitrogen reacted using the molar mass of nitrogen:
Molar mass of nitrogen (N2) = 28.02 g/mol
Mass of nitrogen reacted = Number of moles of nitrogen * Molar mass of nitrogen
= 1.125 mol * 28.02 g/mol
= 31.523 g
Therefore, 31.523 grams of nitrogen reacted in the given reaction.
There is something inconsistent about your numbers. You need more than 18 g of O2 to make 56 g of NO2. The stoichiometry does not match up.
Check your numbers again.