Answer : The time taken for the reaction is, 28 s.
Explanation :
Expression for rate law for first order kinetics is given by :
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 0.0632
t = time taken for the process = ?
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount or concentration of the reactant = 1.28 M
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= amount or concentration left time 't' = 
Now put all the given values in above equation, we get:
Therefore, the time taken for the reaction is, 28 s.
21.4 g Al * (1 mol / 26.98 g ) * (2 mol Fe / 2 mol Al) = 0.793 mol Fe
91.3 g Fe2O3 * (1 mol / 159.69 g) * (2 mol Fe / 1 mol Fe2O3) = 1.14 mol Fe
0.793 mol Fe * (55.85 g / 1 mol) = 44.3 g Fe produced.
Answer:
25%
Explanation:
Ammonium Chloride NH4CI Fertilizer-grade NH4Cl usually contains 25% N. About two-thirds of the world capacity for manufacture of this material is located in Japan, with the remaining one-third situated in India and China.
Answer: Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.
Explanation: Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
