Calculate the mass of 45.0 L of NH3 at 57.0° C and 900. mm Hg. Use the Ideal Gas Law formula and here are values for R:

8.134 (L * kPa)/(mol * K)

0.0821 (atm * L)/(mol * K)

62.364 (L * mmHg)/(mol * K)

Include the following with your

Which Gas Law constant did you use? Why?
The numerical answer to the question.
An explanation of the correct number of significant figures you will use for the numerical answer.

Mass = 33.515 g

Explanation:

Given data:

Volume of ammonia = 45.0 L

Temperature = 57.0°C (57 +273 = 330 K)

Pressure = 900 mmHg

Mass of ammonia = ?

Solution:

Formula:

PV = nRT

We will use R = 62.364 (L * mmHg)/(mol * K) because pressure is given in mmHg.

900 mmHg × 45.0 L = n × 62.364L * mmHg/mol * K × 330 K

40500 mmHg.L = n ×  20580.12L * mmHg/mol

n= 40500 mmHg.L/ 20580.12L * mmHg/mol

n = 1.9679 mol

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 1.9679 mol × 17.031 g/mol

Mass = 33.515 g

When we multiply or divide the values the number of significant figures must be equal to the less number of significant figures in given value.

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .

catya your not at all which answer i've seen you do this on another question!