The answer is TRUE.
If the Energy is on the left, then the problem is true. If it is on the right then it would be negative, false, and considered as exothermic.
Endothermic reaction = the products are higher in energy than the reactants.
Exothermic reaction = a chemical reaction that releases energy by light or heat.
The correct answer would be 2
I’m just letting you know this is really easy you just calculate the molar mass of each compound and divide the amount of the compound (grams) by the molecular Mass
The grams of potassium chlorate that are required to produce 160 g of oxygen is 408.29 grams
<u><em>calculation</em></u>
2 KClO₃→ 2 KCl + 3O₂
Step 1: find the moles of O₂
moles = mass÷ molar mass
from periodic table the molar mass of O₂ = 16 x2 = 32 g/mol
moles = 160 g÷ 32 g/mol = 5 moles
Step2 : use the mole ratio to determine the moles of KClO₃
from equation given KClO₃ : O₂ is 2:3
therefore the v moles of KClO₃ = 5 moles x 2/3 = 3.333 moles
Step 3: find the mass of KClO₃
mass= moles x molar mass
from periodic table the molar mass of KClO₃
= 39 + 35.5 + (16 x3) =122.5 g/mol
mass = 3.333 moles x 122.5 g/mol =408.29 grams
Answer:
The percent yield of NaCl is 78.7 %
Explanation:
CuCl₂ + 2NaNO₃ → Cu(NO₃)₂ + 2NaCl
If the NaNO₃ is determined to be in excess, the limiting reagent is the chloride. We convert the mass to moles:
31 g . 1mol / 134.45g = 0.230 moles
Ratio is 1:2, so we can make a rule of three to determine the theoretical yield
1 mol of copper (II) chloride reacts to produce 2 moles of sodium chloride
Then, 0.230 moles of CuCl₂ will react to produce (0.230 .2) /1 ) = 0.461 moles of NaCl → we convert the moles to mass → 0.461 mol . 58.45 g / 1mol = 26.9 g
To find percent yield we do → (Yield produced / Theoretical yield) . 100
(21.2 g / 26.9 g) . 100 = 78.7 %
