<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.
V = nRT/P
V = 0.685 mol*(.0821 L*atm/K*mol)*273 K/1 atm
The energy of the carbide released is 7262.5MJ.
<h3>What is the energy?</h3>
We know that the reaction between calcium oxide and carbon occurs in accordance with the reaction; 
. The reaction is seen to produce 464.8kJ of energy per mole of carbide produced.
Number of moles of 
produced = 1000 * 10^3 g/64 g/mol
= 15625 moles of calcium carbide
If 1 mole of transfers 464.8 * 10^3 J
15625 moles of calcium carbide transfers 15625 moles * 464.8 * 10^3 J/ 1 mol
= 7262.5MJ
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The equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.
<h3>What is the equilibrium expression?</h3>
The equilibrium expression shows how much of the reactants are converted into products. If the equilibrium constant is large and positive, most of the reactants have been converted into products.
Thus, the equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.
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