2NH3+ 2Na---) 2NaNH2+1H2
assuming that you start with 25.7 g of ammonia gas and 22.4 g of sodium metal and asuming that the reaction goes to completion what is the mass in grams of NaNH2
Find the limiting reagent.
25.7g of NH3*(1 mole/17.031 g)= moles of NH3
22.4g of Na *(1 mole/22.990 g)= moles of Na
Since moles of Na= moles of NH3=moles of NaNH2
The lowest number of moles from the above calculations multiplied by molecular weight will give you the amount of NaNH2
moles of limiting reagent*(39.01 g/mol)= mass of NaNH2
To determine the mass of NaNH2 produced in the reaction, we need to find the limiting reactant between ammonia gas (NH3) and sodium metal (Na).
1. First, we need to determine the number of moles for each reactant using their respective molar masses:
The molar mass of NH3 (ammonia) is 17.03 g/mol.
The molar mass of Na (sodium) is 22.99 g/mol.
Moles of NH3 = mass of NH3 / molar mass of NH3
Moles of NH3 = 25.7 g / 17.03 g/mol
Moles of NH3 ≈ 1.51 mol
Moles of Na = mass of Na / molar mass of Na
Moles of Na = 22.4 g / 22.99 g/mol
Moles of Na ≈ 0.97 mol
2. Next, we need to determine the stoichiometric ratio between NH3 and NaNH2 using the balanced chemical equation:
2NH3 + 2Na → 2NaNH2 + H2
According to the equation, 2 moles of NaNH2 are produced for every 2 moles of NH3.
Therefore, the moles of NaNH2 produced will be the same as the moles of NH3.
3. Since the reaction goes to completion, and moles are directly proportional to mass, we can conclude that the mass of NaNH2 will be the same as the mass of NH3.
Mass of NaNH2 ≈ Moles of NH3 ≈ 25.7 g
So, the mass of NaNH2 produced in the reaction, assuming it goes to completion, is approximately 25.7 grams.