Answer :
By 13:00, 5.1 mEq of KCl will be left in the remaining 340 mL of the IV solution which is calculated using the total mEq, determining fluid infused, and finding the remaining concentration.
To determine how many mEq of potassium chloride (KCl) will be left in the IV solution at 1300, follow these steps:
- Calculate total mEq of KCl in the whole solution:
The IV solution contains 1000 mL with a concentration of 1.5 mEq/100 mL.
Therefore, the total mEq in 1000 mL is 1.5 mEq/100 mL × 1000 mL = 15 mEq.
- Determine the amount of fluid infused by 1300:
The infusion starts at 0700 and runs until 1300, which is 6 hours.
At an infusion rate of 110 mL/hr, the total volume infused by 1300 is 110 mL/hr × 6 hr = 660 mL.
- Calculate the volume remaining in the IV bag:
With 1000 mL initially, and 660 mL infused, the volume remaining is 1000 mL - 660 mL = 340 mL.
- Determine mEq of KCl left in the remaining solution:
The concentration remains unchanged as infusion doesn't change concentration. Thus, the remaining mEq in the 340 mL is (1.5 mEq/100 mL) * 340 mL = 5.1 mEq.
Therefore, at 1300, there will be 5.1 mEq of KCl left in the remaining 340 mL of fluid.
Answer:
The amount of potassium chloride contained in the remaining intravenous would be 5.1 mEq
Explanation:
Given
that 1000 mL of Dextrose in water contains 1.5 mEq of potassium chloride per 100 mL
Calculating the amount of potassium contained per mL
If 1.5 mEq of potassium chloride is contained in 100 mL of Dextrose in water, 1 mL would be;
per mL = 1.5 m Eq / 100 mL
= 0.015 mEq /mL
So the amount of potassium chloride per mL is 0.015 mEq /mL
Calculating the amount of liquid remaining.
Given the start time = 0700 Hrs
the end time = 1300 Hrs
the time the intravenous lasted = 1300 Hrs - 0700 Hrs = 6 Hrs
So the intravenous lasted for 6 hours.
Therefore at an infusion rate of 110 mL/hr the amount of intravenous infused would be;
6 hrs x 110 mL/hr = 660 mL
Therefore the amount of intravenous infused is 660 mL
The intravenous that remained would be;
1000 mL - 600 mL = 340 mL
The intravenous that would remain after 6 Hours would be 340 mL
Calculating the amount of potassium chloride contained in the remaining intravenous;
Given that the amount of potassium chloride contained in 1 mL of the intravenous is 0.015 mEq;
The amount of potassium chloride contained in the intravenous remaining would be;
0.015 mEq/mL x 340 mL = 5.1 mEq
Therefore the amount of potassium chloride contained in the remaining intravenous would be 5.1 mEq
