Use a ratio to find out that x/190.2 = 74.5/100
hope this helps
Answer:
It is equal to the number of moles of acid that reacted. When Oxalic acid is your limiting reactant it is the # of moles of oxalic acid used. When NaOH is your limiting reactant it is equal to the number of moles of NaOH used.
Answer:
+ axit
CH2=CH-CH2-COOH,
CH3-CH=CH-COOH (tính cả đồng phân hình học)
CH2=C(CH3)-COOH.
+ este
HCOOCH=CH-CH3 (tính cả đồng phân hình học)
HCOO-CH2-CH=CH2,
HCOOC(CH3)=CH2.
CH3COOCH=CH2
CH2=CH-COOCH3
HCI is the limiting reactant, and 4.3 mol AICI3 can be formed.
Answer : The rate law for the overall reaction is, ![Rate=k[NO]^2[H_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BNO%5D%5E2%5BH_2%5D)
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
As we are given the mechanism for the reaction :
Step 1 :
(slow)
Step 2 :
(fast)
Overall reaction : ![2NO+2H_2\rightarrow N_2+2H_2O](https://tex.z-dn.net/?f=2NO%2B2H_2%5Crightarrow%20N_2%2B2H_2O)
The rate law expression for overall reaction should be in terms of
.
As we know that the slow step is the rate determining step. So,
The slow step reaction is,
![2NO+H_2\rightarrow N_2+H_2O_2](https://tex.z-dn.net/?f=2NO%2BH_2%5Crightarrow%20N_2%2BH_2O_2)
The expression of rate law for this reaction will be,
![Rate=k[NO]^2[H_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BNO%5D%5E2%5BH_2%5D)
Hence, the rate law for the overall reaction is, ![Rate=k[NO]^2[H_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BNO%5D%5E2%5BH_2%5D)