What volume of 0.123 M K3PO4 is required to react with 60 mL of 0.862 M MgCl2 according to the equation
2K3PO4+3MgCl2--> Mg3(PO4)2+6KCl
Answer in units of mL.
The correct answer to this question is 6.69 mL
First you have to convert grams to moles
Then add up the numbers
Finally get to the covalent formula and divide by the atomic number of mercury
And final answer 6.69 mL
It is 6.69 mL
Thanks 6ix9ine, it is actually the correct answer
840.9756098
To determine the volume of 0.123 M K3PO4 required to react with 60 mL of 0.862 M MgCl2, we first need to find the mole ratio between K3PO4 and MgCl2 based on the balanced equation.
From the equation: 2K3PO4 + 3MgCl2 → Mg3(PO4)2 + 6KCl
We can see that the ratio of K3PO4 to MgCl2 is 2:3. This means that for every 2 moles of K3PO4, we need 3 moles of MgCl2.
Now, let's calculate the number of moles of MgCl2 we have:
Molarity (M) = moles (n) / volume (V)
0.862 M = n / 0.060 L
Rearranging the equation:
n (moles of MgCl2) = 0.862 M × 0.060 L = 0.05172 moles
Since the ratio of K3PO4 to MgCl2 is 2:3, we can determine the number of moles of K3PO4 using the ratio:
n (moles of K3PO4) = (2/3) × n (moles of MgCl2)
n (moles of K3PO4) = (2/3) × 0.05172 moles = 0.03448 moles
Now, let's calculate the volume of 0.123 M K3PO4 needed using the molarity equation:
Molarity (M) = moles (n) / volume (V)
0.123 M = 0.03448 moles / V
Rearranging the equation:
V (volume of K3PO4) = 0.03448 moles / 0.123 M = 0.2803 L
Since the volume is given in milliliters (mL):
V (volume of K3PO4) = 0.2803 L × 1000 mL/L = 280.3 mL
Therefore, the volume of 0.123 M K3PO4 required to react with 60 mL of 0.862 M MgCl2 is 280.3 mL.
what is the answer
1. Convert MgCl2 to moles. moles = M x L.
2. Using the coefficients in the balanced equation, convert moles MgCl2 to moles K3PO4.
3. Remembering M = moles/L, you have moles and you have M, calculate L and convert to mL.