Question
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.
2K3PO4+3MgCl2--> Mg3(PO4)2+6KCl
Answer in units of mL.
Answers
DrBob222
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.
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.
Blib
840.9756098
bob
what is the answer
6ix9ine
It is 6.69 mL
bob
Thanks 6ix9ine, it is actually the correct answer
Mr. ShaW
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
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