1870 kg = 1870000 g of ammonia
We first have to find out how many moles of ammonia that is. To do this, we divide the grams of ammonia by the molar mass of ammonia (17.031 grams/mole).
1870000/17.031 = 109799.7769 moles of ammonia
From the balanced equation, we know that 2 moles of ammonia will produce 1 mole of fertilizer. So to find out how many moles of fertilizer 109799.7769 moles of ammonia would produce, we multiply that number by the ratio of (1/2). (We always put what we want to find or figure out in the numerator)
109799.7769 x (1/2) = 54899.88844 moles of fertilizer
To convert this into grams, we must multiply 54899.88844 moles by the molar mass of (NH4)2SO4 which happens to be 132.14 grams/mole.
54899.88844 x 132.14 = 7254471.258 grams of fertilizer
This is just the theoretical yield to calculate the actual yield, we multiply this number by the percentage given in the question.
7254471.258 x 0.865 = 6275117.639 grams of actual fertilizer
If we round to 3 sig figs and convert it into kg, we get:
6280 kg of (NH4)2SO4