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How many kilograms of NH3 are needed to produce 2.90 x 10^5 kg of (NH4)2SO4?

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    first, write the chemical reaction:
    NH3 + H2SO4 -> (NH4)2SO4
    then we balance this:
    2NH3 + H2SO4 -> (NH4)2SO4
    then we find the weight of NH3 and (NH4)2SO4. We can do this by getting each atomic mass of the element in the chemical formula, and add them. It can be found on periodic table of elements.
    NH3 = 17 kg/kmol
    (NH4)2SO4 = 132 kg/kmol
    then, we get the number of moles of the product by dividing the given by the mass:
    2.90*10^5 kg (NH4)2SO4 / 132 kg/kmol = 2196.97 kmol (NH4)2SO4
    then we get the number of moles of NH3 required (stoichiometric ratios from the balanced equation):
    2196.97 kmol (NH4)2SO4 * (2 kmol NH3 / 1 kmol (NH4)2SO4) = 4393.93 kmol NH3
    finally, we multiply this by the mass of NH3 to get the total mass required:
    4393.93 kmol NH3 * (17 kg/kmol) = 74696.97 kg NH3

    hope this helps~ :)

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