2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃
m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol
1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw
2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}
m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}
m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}
m(Fe)=55.847*245*0.805/179.857= 61.24 g
Equation: M1V1 = M2V2
Where M = concentration & V = volume
Step 1: Write down what is given and what you are trying to find
Given: M1 = 6.00M, V1 = 2.49mL, and V2 = 50.0mL
Find: M2
Step 2: Plug in the values into the equation
M1V1 = M2V2
(6.00M)(2.49mL) = (M2)(50.0mL)
Step 3: Isolate the variable (Divide both sides by 50.0mL so M2 is by itself)
(6.00M)(2.49mL) / (50.0mL) = M2
Answer: M2 = 0.30M
*Don't forget sig figs & units!
Both ehhevshahahbsbdvhshs
The answer is C
explanation
Explanation:
when acid reacts with water it form salt and water this reaction is called neutralization reaction