Answer:
The mass we will get from AgCl is 0.839g
Explanation:
Hello!
Let's solve this!
The reaction is:
2AgC2H3O2 + MgCl2 ---> 2AgCl + Mg (C2H3O2) 2
The data we have are:
V1 = 75mL * (1L / 1000L) = 0.075L
M1 = 0.078M
V2 = 55mL * (1L / 1000mL) = 0.055L
M2 = 0.109M
The molarity formula is
M = mol / V
We calculate the moles of each compound:
molAgC2H3O2 = 0.075L * 0.078mol / L = 0.00585mol
molMgCl = 0.055L * 0.109M = 0.005995
We calculate the compound that is in excess, and then use the limiting reagent to calculate the amount of product we will obtain.
Since there are two Cl in MgCl2, this means that we have 0.01199 mol to produce the product, but we will only use 0.00585 mol which is the amount we have of AgC2H3O2. So MgCl is the excess reagent.
Then:
AgCl mass = 0.00585mol * (2mol AgCl / 2molAgC2H3O2) * (143.34g AgCl / 1mo AgCl) = 0.839gAgCl
The mass we will get from AgCl is 0.839g
If it is 60 Celsius that would conver to fare height by means of this equation; (1.8*60)+32°F
Which would come out to.... 140° Fahrenheit... Hardly seems like chilly conditions.
The answer would be KMnO4, please let me know if you would like me to explain how i got this