Answer:
Pb(NO3)2 + KI = KNO3 + PbI2
Answer: The reaction order with respect to A is m
Explanation:
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
For the given reaction:
![Rate=k[A]^m[B]^n](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5Em%5BB%5D%5En)
In this equation, the order with respect to each reactant is not equal to its stoichiometric coefficient which is represented in the balanced chemical reaction.
Hence, this is not considered as an elementary reaction.
Order with respect to A = m
Order with respect to B = n
Overall order = m+n
Thus order with respect to A is m.
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant

3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO

From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂

4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂

(b) Mass of NO reacted

(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO