There are 50.0 grams of NH3 + Na which produce NaNH2 + H2 (unbalanced)
How much grams is produced with the excess of the reactant?
3 answers
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.
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.
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.