Show how to get the answer on each of the following problems: Use the following
ID: 860760 • Letter: S
Question
Show how to get the answer on each of the following problems:
Use the following reaction for the problems below:
2Na + Cl2 -> 2NaCl
How many grams of product can be formed from 25.0 g of nonmetal reactant (assuming excess metal reactant is present)? Show your work.
Answer: 41.2 g product
Given 34.7 g of metal reactant and 25.0 g of nonmetal reactant, which is limiting reactant and how much of the excess reactant remains? Show your work.
Answer: Nonmetal is limiting, 16.4 g of metal remain
Given 25 g of each reactant, what is the theoretical yield (what is the maximum amount of product that can be formed under these conditions)? Show your work.
Answer: 41.2 g product
Show how to get the answers given above!
Explanation / Answer
a) since the number of moles of the NaCl = number of moles of the Cl2
wt/58.5 = 25/35.5
wt = (25/35.5)*58.5
wt = 41.19 g = 41.2 grams
b) To find the limiting reactant first we should calculate the number of moles of both the reactants
moles of Na = 34.7/23
= 1.508
moles of the Chlorine = 25/35.5 = 0.704 moles
1.508 moles of Na needed 1.508/2 moles of Cl = 0.754 moles of Cl
but here moles of Cl is less than that of 0.754
non metal Cl is limiting reagent
and the number of moles of Na needed now = 2*0.704 = 1.408
therefore the remaining metal = (24.99*23)/35.5 = 16.4 g of metal
c) since 23 g of Na needed 35.5 gms of Cl
For given both 25 g of the reactants Cl is the limiting reagent
Therefore product wt should be found by considering the Cl reactant
moles of Cl = moles of NaCl
Wt of the product NaCl = (25/35.5)*58.5 = 41.2 grams