How many grams of NH3 can be produced if 105g of NH4OH are completey decomposed?
To determine the number of grams of NH3 that can be produced when 105g of NH4OH is completely decomposed, we need to consider the chemical equation for the decomposition reaction.
The chemical equation for the decomposition of NH4OH (ammonium hydroxide) is:
NH4OH → NH3 + H2O
From the equation, we can see that 1 mole of NH4OH will produce 1 mole of NH3.
To calculate the number of moles of NH4OH in 105g, we need to use its molar mass. The molar mass of NH4OH can be calculated by summing the atomic masses of nitrogen (N), hydrogen (H), and oxygen (O).
Molar mass of NH4OH:
(1x N) + (4x H) + (1x O) = 14.01 + 4.03 + 16.00 = 34.04 g/mol
Now, we can calculate the number of moles of NH4OH in 105g using the formula:
Number of moles = Mass (g) / Molar mass (g/mol)
Number of moles of NH4OH = 105g / 34.04 g/mol = 3.08 mol
Since the molar ratio between NH4OH and NH3 is 1:1, we can conclude that 3.08 moles of NH4OH will produce 3.08 moles of NH3.
Finally, we can calculate the mass of NH3 produced by multiplying the number of moles of NH3 by its molar mass.
Molar mass of NH3:
(1x N) + (3x H) = 14.01 + 3.03 = 17.04 g/mol
Mass of NH3 = Number of moles of NH3 x Molar mass of NH3
Mass of NH3 = 3.08 mol x 17.04 g/mol = 52.51 g
Therefore, when 105g of NH4OH is completely decomposed, approximately 52.51 grams of NH3 can be produced.