Answer:
moles of ammonia produced = 0.28 moles
Explanation:
The reaction is
As per equation, one mole of nitrogen will react with three moles of hydrogen to give two moles of ammonia
So 0.140 moles of nitrogen will react with = 3 X 0.140 moles of Hydrogen
= 0.42 moles of hydrogen molecule.
this will give 2 X 0.140 moles of ammonia = 0.28 moles of ammonia
the moles of ammonia produced = 0.28 moles
Here the nitrogen is limiting reagent.
The question is incomplete, here is the complete question.
A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.
<u>Answer:</u> The concentration of copper fluoride in the solution is
<u>Explanation:</u>
To calculate the molarity of solute, we use the equation:
We are given:
Given mass of copper (II) fluoride = 0.0498 g
Molar mass of copper (II) fluoride = 101.54 g/mol
Volume of solution = 100.0 mL
Putting values in above equation, we get:
Hence, the concentration of copper fluoride in the solution is