What volume of 0.123 M K3PO4 is required to react with 60 mL of 0.862 M MgCl2 according to the equation

Question

2K3PO4+3MgCl2--> Mg3(PO4)2+6KCl

Answer in units of mL.

Answer 1

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.

Answer 2

840.9756098

Answer 3

what is the answer

Answer 4

It is 6.69 mL

Answer 5

Thanks 6ix9ine, it is actually the correct answer

Answer 6

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