Answer:
3.785 L of lye solution needed for 907.185 g of fat
Explanation:
The computation of the number of grams in 2.00 pounds is shown below:
As we know that
One gallon of lye solution needed 2 fats pounds
Also,
1 gallon = 3.785 L
And,
1 lb = 453.592 g
So,
For 2 lb, it is
= 2 × 453.592
= 907.185 g
Therefore
3.785 L of lye solution needed for 907.185 g of fat
hence, the same is to be considered