3. Given the following equation:
1 answer:
Answer:
4.767 grams of KCl are produced from 2.50 g of K and excess Cl2
Explanation:
The balanced equation is
2 K+ Cl2 --->2 KCI
Here the limiting agent is K. Hence, the amount of KCl will be calculated as per the mass of 2.50 gram of K
Mass of one atom/mole of potassium is 39.098 grams
Number of moles is 2.5 grams =
So, 2 moles of K produces 2 moles of KCL
0.064 moles of K will produces 0.064 moles of KCl
Mass of one molecule of KCl is 74.5513 g/mol
Mass of 0.064 moles of KCl is 4.767 grams
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Answer:
a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)If concentration of is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d) If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
Explanation:
Sucrose + fructose+ glucose
The rate law of the reaction is given as:
[sucrose]= 1.0 M
..[1]
a)
The rate of the reaction when [Sucrose] is changed to 2.5 M = R'
..[2]
[2] ÷ [1]
If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)
The rate of the reaction when [Sucrose] is changed to 0.5 M = R'
..[2]
[2] ÷ [1]
If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)
The rate of the reaction when is changed to 0.001 M = R'
..[2]
[2] ÷ [1]
If concentration of is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d)
The rate of the reaction when [sucrose] and both are changed to 0.1 M = R'
..[2]
[2] ÷ [1]
If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.