There are 50.0 grams of NH3 + Na which produce NaNH2 + H2 (unbalanced)
How much grams is produced with the excess of the reactant?
You must re-write your question and make it clear. 50 grams of which reactant. Which reactant is the excess.
Both of the reactants are 50.0 grams and NH3 is the amount that has an excess.
Thanks.
2NH3 + 2Na ==> 2NaNH2 + H2
Step 1. Write and balance the equation, as above.
2. I will do this for both reactants and you can take your pick.
a. Convert grams NH3 to moles. moles = grams/molar mass.
b. Convert grams Na to moles. moles = grams/molar mass.
3a. Using the coefficients in the balanced equation convert moles NH3 from 2a to moles NaNH2 (and separately to moles H2 if you need that, too).
3b. Same procedure, convert moles Na in 2b to moles NaNH2 (and separately to moles H2 if you need it.)
3c. It is likely that the answers from 2a and 2b (for the same product) will not be the same; obviously, one of them is wrong. The correct value is ALWAYS the smaller one and the reactant producing the smaller value is the limiting regent.
4. You can convert any of the products from 3 to grams. grams = moles x molar mass.
To determine how many grams are produced with the excess reactant, you first need to balance the chemical equation. The given unbalanced equation is:
NH3 + Na -> NaNH2 + H2
Now, we need to balance the equation by adding coefficients to each compound until the numbers of atoms on both sides of the equation are the same. After balancing, the equation becomes:
2NH3 + 2Na -> 2NaNH2 + H2
The coefficient "2" in front of NH3, Na, and NaNH2 indicates that two moles of each compound are involved in the reaction. However, only one mole of H2 is produced.
Since we have 50.0 grams of NH3 + Na, we need to determine the limiting reactant to determine the amount of NaNH2 + H2 produced.
To do this, we need to compare the molar masses of NH3 and Na and determine which one produces fewer moles. The molar mass of NH3 is approximately 17.03 grams/mol, and the molar mass of Na is approximately 22.99 grams/mol.
Now we can calculate the number of moles of NH3 and Na:
Moles of NH3 = (mass of NH3 in grams) / (molar mass of NH3)
= (50.0 g) / (17.03 g/mol)
≈ 2.94 mol
Moles of Na = (mass of Na in grams) / (molar mass of Na)
= (50.0 g) / (22.99 g/mol)
≈ 2.17 mol
Since Na produces fewer moles, it is the limiting reactant, meaning that all of the Na will be consumed in the reaction. Thus, 2.17 moles of NaNH2 and 2.17 moles of H2 will be produced.
To determine the grams of NaNH2 + H2 produced, we can multiply the number of moles by their respective molar masses:
Grams of NaNH2 = (moles of NaNH2) * (molar mass of NaNH2)
= (2.17 mol) * (39.0 g/mol)
≈ 84.63 g
Grams of H2 = (moles of H2) * (molar mass of H2)
= (2.17 mol) * (2.02 g/mol)
≈ 4.39 g
Therefore, with the excess reactant, approximately 84.63 grams of NaNH2 and 4.39 grams of H2 will be produced.