I believe the answer is a balance or mechanical scale
<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
Answer:
12 moles of H₂O are formed in this combustion.
Explanation:
First of all, think the reaction:
2CH₃OH (l) + 3O₂ (g) → 2CO₂ (g) + 4H₂O (g)
Ratio in the reactants is 2:3, so 2 mol of methanol need 3 mol of oxygen to react. Then 8 mol of CH₃OH, will need (8.3)/2 = 12 moles of O₂
We have 9 moles of O₂, so this is the limiting reactant.
3 mol of oxygen produce 4 mol of water
Then, 9 mol of oxygen will produce ( 9 .4)/3 = 12 moles
Looks correct but the second to last I would of put abiotic and biotic factors but I don’t know what’s right for you
