posted by Diana on .
How many milliliters of NH3 (at STP) are needed to react with 45.0 mL of NO2 (at STP) according to the equation.
Can someone please... help me?
8NO2 (g) + 6NH3 (g) --> 7N2) (g) + 9H2O (g)
IS this help?
a) 45.0 mL
b) 27.5 mL
c) 33.8 mL
d) 60.0 mL
I got C). 33.75 or 33.8. Use Spectrometry. You have 45.0ml of NO2 right. Convert that to Liters so you get .045L. Since its at STP convert .045L to moles using 22.4 L/mol.
(.002 moles of NO2)[(6 moles of NH3)/(8 moles of NO2)] = .0015 mole of NH3
use 22.4 L/mol to convert it to L and multiply it by 1000 to convert to milliliters.
Hope that helps.
Amphee is correct but let me introduce both of you to a shortcut that can be used WHEN ALL OF THE MATERIALS ARE GASES.
The rationale first:
Since you divide by 22.4 to convert to moles, then multiply by 22.4 to convert back to volume, you can omit both of those steps and simply use the coefficients as the following:
45 mL NO x (6 mols NH3/8 moles NO) = 33.75 mL NH3.
Thank you so much both of you. It really help me to understand. :)