Answer:
0.1388 g
Explanation:
The mass of obtained on precipitation = 236 mg
1 mg = 0.001 g
Thus, Mass of = 0.236 g
Molar mass of = 233.43 g/mol
The formula for the calculation of moles is shown below:
Thus,
Moles of = 0.001011 moles
According to the reaction,
Thus, moles of barium = 0.001011 moles
Molar mass of barium = 137.327 g/mol
Thus, Mass = Moles * Molar mass = 0.001011*137.327 g = 0.1388 g
Answer:
The amount of NO₂ that can be produced 8.533 g
Explanation:
According to question
2 NO(g) + O₂(g) → 2 NO₂(g)
Given
Moles of nitrogen monoxide = 0.377
Moles of oxygen = 0.278
Since 'NO' is the limiting reagent according to this ratio.
According to equation
2 moles NO reacts to form 2 moles NO₂
So, 0.1855 moles NO give = 0.1855 moles of NO₂
Mass of 1 mole NO₂ = 46 g/mole
Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g