Basic Chem - Molar Percentage problem.

  • Thread starter Thread starter MarcMTL
  • Start date Start date
  • Tags Tags
    Chem
AI Thread Summary
To calculate the molar percentage of nitrogen in Ni(NO2)2, the formula used is mol% of A = n(A) / n(total) * 100. The confusion arises from the 1 kg mass of the compound, but molar percent is indeed independent of mass. The correct approach involves determining the number of moles of nitrogen relative to the total moles in the compound. The molar percent of nitrogen can be calculated using the ratio of nitrogen moles to the total moles of the compound. Therefore, the mass does not need to be considered for this calculation.
MarcMTL
Messages
25
Reaction score
0

Homework Statement



Calculate the molar percent of nitrogen in 1kg of Ni(NO2)2.

Homework Equations


mol% of A = n(A) / n(total) * 100

The Attempt at a Solution



I'm stuck on this basic question. I have a tendency to say that the molar percent of nitrogen would basically be \frac{2}{1+2+4} * 100, basically the nb of moles of N over the sum of the total.

What's causing me confusing is the 1kg.

My question is: if I were looking for the specific number of moles, then the mass would be accounted for. But if it's molar percentt that's wanted, wouldn't it be independent of the mass?

Thanks.

Marc
 
Physics news on Phys.org
Yes, molar percent is mass independent. You can safely ignore 1 kg.

--
methods
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top