Potassium phosphate vials contain 3 mmol of phosphate and 4.4 mEq of potassium per mL, and potassium chloride vials contain 2 mEq of potassium per mL.

How many milliliters of potassium phosphate and how many milliliters of potassium chloride are required to make LB's 2-in-1 parenteral nutrition order with:

- Potassium = 65 mEq
- Phosphate = 24 mmol

To calculate the amount of potassium phosphate and potassium chloride required to make the 2-in-1 parenteral nutrition order, we need to calculate the number of milliliters of each solution needed. Let's start with potassium phosphate:

Given that the potassium requirement is 65 mEq and the concentration of potassium in the potassium phosphate vials is 4.4 mEq/mL, we can calculate:

Potassium phosphate mL = (required potassium mEq) / (potassium concentration mEq/mL) = 65 mEq / 4.4 mEq/mL = 14.77 mL

Next, let's calculate the amount of potassium chloride needed:

Given that the potassium requirement is 65 mEq and the concentration of potassium in the potassium chloride vials is 2 mEq/mL, we can calculate:

Potassium chloride mL = (required potassium mEq) / (potassium concentration mEq/mL) = 65 mEq / 2 mEq/mL = 32.5 mL

Learn more about potassium chloride here:

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