If you combine 2.00 ml 2.0 x 10^-3 M KSCN and 98.00 ml 2.0M Fe(NO3)3 what is the concentration of FeSCN?
I know the equation (V concentrated*M concentrated)=(V diluted *M diluted) should be used, but how do you go about that when you have two initial substances?
To find the concentration of FeSCN, we can use the equation you mentioned: (V concentrated * M concentrated) = (V diluted * M diluted). In this case, we have two initial substances: KSCN and Fe(NO3)3.
Let's break down the problem step by step:
1. Calculate the moles of KSCN and Fe(NO3)3.
- Moles = Volume (in liters) × Concentration (in Molarity)
Moles of KSCN = 2.00 ml × (2.0 x 10^-3 M) = 4.00 x 10^-6 moles
Moles of Fe(NO3)3 = 98.00 ml × (2.0 M) = 1.96 moles
2. Identify the limiting reactant.
- The limiting reactant is the one that is present in a lower amount and determines the amount of product formed. In this case, KSCN is the limiting reactant as it is present in a much smaller amount.
3. Use stoichiometry to determine the moles of FeSCN formed.
- From the balanced chemical equation, we know that the ratio between FeSCN and KSCN is 1:1.
Moles of FeSCN = Moles of KSCN = 4.00 x 10^-6 moles
4. Calculate the final volume of the diluted solution.
- The final volume is the sum of the volumes of the two initial solutions.
Final volume = 2.00 ml + 98.00 ml = 100.00 ml = 0.100 L
5. Calculate the concentration of FeSCN.
- Using the equation (V concentrated * M concentrated) = (V diluted * M diluted), we can rearrange it to solve for the unknown concentration.
(0.100 L) * (M FeSCN) = (0.100 L) * (Moles of FeSCN)
M FeSCN = (Moles of FeSCN) / (0.100 L)
M FeSCN = (4.00 x 10^-6 moles) / (0.100 L)
M FeSCN = 4.00 x 10^-5 M
Therefore, the concentration of FeSCN is 4.00 x 10^-5 M.