According to the balanced reaction below, calculate the moles of NO2 that form when .00520 mole of N2O5 completely reacts: N₂O₅(g) → NO₂(g) + O₂(g)?

Answer 1

2N₂O₅ ==> 4NO₂ + O₂ ... balanced equation

0.00520 moles N₂O₅ x 4 moles NO₂ / 2 moles N₂O₅ = 0.0104 moles NO₂ are formed

Answer 2

1. Balance the equation:( the coefficients in balanced equation equals number of moles)

2 N₂O₅(g) _----> 4 NO₂(g) ₊ O₂(g)

_{2 moles of dinitrogen pentoxide gas decomposes into 4 moles of nitrogen dioxide gas and 1 mole of oxygen gas}

1. Make a plan(this is a mole to mole problem)

moles A(given substance) to moles B(wanted substance)

1. write down given moles and substance and multiply by the correct conversion factor( since you are converting mol to mol, multiply by the mole ratio between the given and wanted substances

mol given substance x (mol wanted/ mol given) = mol wanted substance

0.00520 mol N₂O_{5 X} (4 mol NO₂/ 2 mol N₂O₅) = 0.0104 mol NO₂

I want to thank J.R.S for catching my mistake:)

Answer 3

Aha! The question is misleading because it begins, "According to the balanced reaction below," but in fact the reaction below that is NOT balanced! In order to proceed, we first need to balance it.

DiNitrogen pentaoxide deomposes: N₂O₅ --> NO₂ + O₂ Now straight away we perceive that this equation has two nitrogens on the left (reactant side) but only one nitrogen on the right. So let's put a 2 in front of NO₂

N₂O₅ --> 2NO₂ + O₂

Now the equation has the same number of nitrogens (one times two and two times one) on each side. It also has the same number of oxygens (5) on each side. That is, the equation is balanced. Now we can proceed.

0.00520 moles of N₂O₅ times 2 moles of NO₂ per mole of N₂O₅ = 0.0104 moles of N₂O₅ . That is the answer.

Note that we must not write 0.01040 moles of N₂O₅ because that would have four significant figures and we are only entitled to three.