Comparing %N and %S by mass in fertilizers X and Y

Steps:

• Calculate %N in NH4NO3: (2×14)/80 = 35%; in (NH4)2SO4: (2×14)/132 ≈ 21.2%.
• For X (500 g each): total N = (500×0.35) + (500×0.212) ≈ 281 g (28.1%); %S from (NH4)2SO4 = (500×32/132) ≈ 12.1%.
• For Y: %N from NH4NO3 = 24.5%; %S from CaSO4 = (300×32/136) ≈ 7.1%.
• X has higher %N (28.1% > 24.5%) and higher %S (12.1% > 7.1%).

Why A is correct:

• Calculations show X provides more N from dual sources and more S from sulfate, exceeding Y's single-source N and lower S content.

Why the others are wrong:

• B: Y has lower %N than X.
• C: X has higher %S than Y.
• D: X has higher %N than Y.

Final answer: A