# 3H2 + N2 => 2NH3 determine the mass of ammonia produced if 3.50 mol reacts with 5.00 mol N2

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3H2 + N2 => 2NH3 determine the mass of ammonia produced if 3.50 mol reacts with 5.00 mol N2

So, since we already have the balanced equation, we can go ahead and convert what we know, to what we want. So, since one of the reactants (hydrogen gas or nitrogen gas) will be limiting the reaction, we have to convert both of them to moles of NH3 in order to figure out how much NH3 will actually be produced (It’ll be the smaller number)

so we take 3.50moles of H2 and multiply it by the mole to mole ratio (so the coefficients of what’s in front), with moles of H2 on the bottom

so now we have 3.50moles of H2 multiplied by 2/3. which will give us about 2.33 moles of NH3

now we do the same for N2:

5.00moles of N2 multiplied by 2/1 (remember the ratio, the bottom number is the coefficient of N2) which gives us 10moles of NH3

so we know that H2 is limiting, so we take the 2.33moles of NH3 that is formed, and convert it into grams (since it wants mass)

so all we do is multiply the 2.33moles and its molar mass:

2.33 x 17.03 = 39.68g of NH3 produced