Answer:
% of N present in 12.2 oz of fertlilizer = 51.9 g
Step-by-step explanation:
Given:
% of N in fertilizer = 15
Mass of sample fertilizer = 12.2 oz
Conversion factor for Ounce to gram:
1 ounce = 28.3495 g
12.2 ounce (oz) = 12.2 × 28.3495
= 345.8639 g
% of N in fertlilizer = 345.8639 × 0.15
= 51.8795 g
= 51.9 g
% of N present in 12.2 oz of fertlilizer = 51.9 g