Calculate the volume of ammonia that is released when 890 g of ammonium nitrate, containing 20%
Calculate the volume of ammonia that is released when 890 g of ammonium nitrate, containing 20% impurities, interacts with sodium hydroxide solution. how many grams of a 20% alkali solution will it take for the reaction?
This reaction proceeds according to the following chemical reaction equation:
NH4NO3 + NaOH = NH3 + NaNO3;
Find the chemical amount of ammonium nitrate. To do this, we divide its weight by its molar mass.
M NH4NO3 = 14 + 4 + 14 + 16 x 3 = 80 grams / mol;
N NH4NO3 = 890 x 0.8 / 80 = 8.9 mol;
During the reaction, the same volume of ammonia will be released.
Let’s define its volume.
To this end, we multiply the amount of substance by the standard volume of 1 mole of gas (which is 22.40 liters)
V NH3 = 8.9 x 22.4 = 199.36 liters;
Calculate the weight of 8.9 mol of sodium hydroxide.
M NaOH = 23 + 16 +1 = 40 grams / mol;
m NaOH = 8.9 x 40 = 356 grams;
The weight of a 20% solution will be: 356 / 0.2 = 1,780 grams;