How do you calculate mass using ideal gas equation?

Question

Hello everyone, I am having some trouble with my homework.

"From the volume, temperature, and pressure data given, calculate the number of moles and the mass in grams for each gas listed using the ideal gas equation.

a. 2000.0 cm^3 NH3 at 10.0 degrees C and 105.0 kPa"

Answer 1

PV = nRT n = PV/RT P=105.0kPa V=2.000L T=283K mol.wt.NH3=17g/mol R=8.3145 L kPa/K mol

n = (105.0)(2.000)/(8.8.3145)(283) = 0.089mol NH3

0.089molNH3 x 17gNH3/1molNH3 = 1.5g NH3

Answer 2

You use PV=nRT in the usual way to calculate n, the number of moles. Then multiply this number by the molar mass of NH3 (17 mol/g) to get the number of grams.

Another thing to be careful of here is to make sure you put everything in the correct units. Assuming that the R value you're using is the usual one in (L atm/mol K), make sure you convert your volume to L from cubic centimeters, your temperature from Celsius to Kelvin, and the pressure from kPa to atm. (All these conversion factors should be in your textbook.)

Answer 3

Methane is a gasoline a million.0kg = 1000g Moles = mass/Mr one thousand/sixteen =sixty two.5 moles it quite is desirable :) i haven't been taught a thank you to do your 2nd question, (i'm doing AS-point chemistry too!) reliable success!

Answer 4

PV=nRT
n=PV/RT
V=2000.0cm^3=2.00dm^3=2.00L
T=10.0C + 273K= 283K
P=105.0kPa
R=8.314JK^-1mol^-1


n(NH3)=(105.0kPa x2.00L)/8.314JK^-1mol^-1 x 283K=0.0892mol
N2 + 2H3 ===> 2NH3

PV=(mRT)/MM
PV=gRT/MM
g=(PV x MM)/RT

g(N2)=(105.0kPa x 2.00L x 28.0g/mol)/8.314JK^-1mol^-1 x283K=2.500g

g(H2)= (105.0kPa x 2.00L x6.06g/mol)/8.314JK^-1mol^-1 x283K= 0.52087g