A= 6
B= 2
C= 1
D= 5
E= 3
F= 4
The theoretical yield is 204.4 g while the percent yield is 2.57%.
<h3>What is theoretical yield?</h3>
Theoretical yield is the amount of product obtained based on the stoichiometry of the reaction.
S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)
Number of moles of sulfur = 3.25 g /8(32) = 0.013 moles
Number of moles of sodium sulfite = 13.1 g/126 g/mol = 0.103 moles
Since 1 moles of sulfur reacts with 8 moles of sodium sulfite
0.013 moles reacts with 0.013 moles × 8 moles /1 mole = 0.104 moles
There is not enough sodium sulfite hence it is the limiting reactant.
1 mole of sodium sulfite yields 8 moles of product
0.103 moles of sodium sulfite yields 0.103 moles × 8 moles /1 mole = 0.824 moles
Mass of product = 0.824 moles × 248 g/mol = 204.4 g
percent yield = 5.26 g /204.4 g × 100/1
= 2.57%
Learn more about percent yield: brainly.com/question/2506978
Answer:
Explanation:
There is a formula for this:
M = DRT/P where M = molar mass. This just derived from PV = nRT where you say n = grams/molar mass. However, just with this formula, we can get D which is density at STP (1 atm and 273K). We find that D = 6.52g/L.
yes substances Do react by mass
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
