How many grams of N2F4 can be produced from 225gF2?

Answer 1

I love the smell of stoichiometry in the morning! The answer is approximately 300g of N2F4, and here's how you get it:

Step 1: Get an equation for this reaction. Your product is obviously N2F4 (dinitrogen tetrafluoride), and one of the two reactants is F2 (diatomic fluorine gas). The other reactant has to be nitrogen, so you set up and balance a synthesis reaction equation:

N2 + 2F2 --> N2F4

Step 2: Get the number of moles of your reactant by setting up a direct proportion with the Atomic Mass of fluorine:

39g/1mol=225g/x, solve for x to get 5.77 moles of fluorine.

Step 3: Work the molar ratio in the reaction equation to see that you need half as many moles of product as you have of fluorine reactant:

5.77/2=2.89 moles N2F4

Step 4: Set up another direct proportion with the molar mass of product, which is 104g/mol:

104/1=x/2.89, solve for x to get 300.5g, so the answer is approx. 300g of N2O4.

Some things to remember in this calculation:

- don't forget your diatomics

- watch your equation balancing

- label all numbers with correct units so you can keep track of them

Answer 2

For every 1 mole of F2 used, 1 mole of N2F4 is produced. The molar mass of F2 is 38.0 g/mol and N2F4 is 104.0 g/mol. Thus, 225g of F2 would produce 225g/38.0g/mol = 5.92 moles of F2, which would then produce 5.92 moles of N2F4. Therefore, 5.92 moles of N2F4 is 617.28 g.