If 23.2 grams of ammonium nitrite react to form 10.1 grams of nitrogen, how many grams of water must simultaneously be formed?
Well, let's do the math...but wait, I thought you said we were going to have a "watery" good time! Okay, let's dive in.
The balanced chemical equation for the reaction of ammonium nitrite (NH4NO2) is 2NH4NO2 ➞ N2 + 2H2O.
So, for every mole of ammonium nitrite, we get 2 moles of water. The molar mass of ammonium nitrite is approximately 64 grams/mole, and the molar mass of water is 18 grams/mole.
Given that we have 23.2 grams of ammonium nitrite, we can start by calculating the number of moles:
23.2 grams of NH4NO2 × (1 mole/64 grams) = 0.3625 moles of NH4NO2
Since the ratio is 1:2 for NH4NO2 to H2O, we can multiply the moles of NH4NO2 by 2 to find the moles of water:
0.3625 moles of NH4NO2 × 2 moles of H2O = 0.725 moles of H2O
Now, we can multiply the moles of water by the molar mass of water to find the grams of water:
0.725 moles of H2O × (18 grams/1 mole) = 13.05 grams of H2O
So, according to my calculations, approximately 13.05 grams of water should be formed. But remember, these calculations are no joking matter...well, maybe a little bit!
To determine the grams of water formed, we first need to balance the given chemical equation for the reaction of ammonium nitrite:
NH4NO2 -> N2 + 2H2O
The molar mass of NH4NO2 is 64.04 g/mol.
Divide the given mass of ammonium nitrite (23.2 g) by its molar mass to find the number of moles:
23.2 g NH4NO2 / 64.04 g/mol = 0.362 mol NH4NO2
According to the balanced equation, for every 1 mole of NH4NO2, 2 moles of water are formed.
So, the number of moles of water formed will be twice the number of moles of ammonium nitrite:
0.362 mol NH4NO2 * 2 = 0.724 mol H2O
Now, we can calculate the mass of water formed using the molar mass of water (18.015 g/mol):
0.724 mol H2O * 18.015 g/mol = 13.05 g H2O
Therefore, approximately 13.05 grams of water must be formed simultaneously.
To determine the number of grams of water formed in the reaction, we need to use the balanced chemical equation for the reaction of ammonium nitrite (NH4NO2) to form nitrogen (N2) and water (H2O).
The balanced chemical equation for the reaction is:
NH4NO2 → N2 + 2H2O
From the equation, we can see that for every 1 mole of ammonium nitrite (NH4NO2) that reacts, 2 moles of water (H2O) are formed.
To find the number of moles of ammonium nitrite, we need to divide the given mass (23.2 grams) by its molar mass.
The molar mass of ammonium nitrite (NH4NO2) can be calculated by summing the molar masses of its constituent elements:
Molar mass of NH4NO2 = (1 x 14.01) + (4 x 1.01) + 14.01 + (2 x 16.00)
= 80.05 g/mol
Now, we can calculate the number of moles of ammonium nitrite:
Number of moles of NH4NO2 = Mass / Molar mass
= 23.2 g / 80.05 g/mol
= 0.2899 mol (approximately)
According to the balanced chemical equation, for every 0.2899 moles of ammonium nitrite that react, 2 moles of water are formed.
Therefore, the number of moles of water formed can be calculated as:
Number of moles of H2O = 2 x Number of moles of NH4NO2
= 2 x 0.2899 mol
= 0.5798 mol (approximately)
Finally, to determine the mass of water formed, we multiply the number of moles of water by its molar mass.
The molar mass of water (H2O) is calculated by summing the molar masses of its constituent elements:
Molar mass of H2O = (2 x 1.01) + 16.00
= 18.02 g/mol
Mass of H2O = Number of moles of H2O × Molar mass of H2O
= 0.5798 mol × 18.02 g/mol
= 10.44 grams (approximately)
Therefore, approximately 10.44 grams of water must be formed simultaneously in the reaction.