The reaction is O3 + NO arrow O2 + NO2
If 0.740 g O3 reacts w/ 0.670g of NO, how many grams of NO2 will be produced?
I've tried to solve this.
but I can't seem to get the right answer
To solve this problem, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the moles of each reactant:
Moles of O3 = mass of O3 / molar mass of O3
= 0.740 g / 48 g/mol (molar mass of O3)
= 0.01542 mol
Moles of NO = mass of NO / molar mass of NO
= 0.670 g / 30 g/mol (molar mass of NO)
= 0.02233 mol
Now let's look at the balanced equation:
O3 + NO → O2 + NO2
The stoichiometric ratio of O3 to NO2 is 1:1, which means that for every 1 mol of O3, 1 mol of NO2 is produced.
Since the stoichiometric ratio of NO to NO2 is also 1:1, for every 1 mol of NO, 1 mol of NO2 is produced.
Comparing the moles of O3 (0.01542 mol) and NO (0.02233 mol), we can see that NO is the limiting reactant because there are fewer moles of NO than O3.
To find the moles of NO2 produced, we can use the stoichiometric ratio between NO and NO2, which is 1:1. Therefore, the moles of NO2 produced will be equal to the moles of NO.
Moles of NO2 = moles of NO
= 0.02233 mol
Finally, let's calculate the mass of NO2 produced:
Mass of NO2 = moles of NO2 * molar mass of NO2
= 0.02233 mol * 46 g/mol (molar mass of NO2)
= 1.026 g
Therefore, 1.026 grams of NO2 will be produced.
Here is an example I have posted to show how to work stoichiometry problems. It isn't your problem but most stoichiometry problems are alike. Here is a hot link.
http://www.jiskha.com/science/chemistry/stoichiometry.html