my question is this: what is the maximum mass of ammonia that can be produced from a mixture of 1.00 x 10^3g N2 and 5.00 x 10^2g H2
the equation given is N2(g) + 3H2(g) -> 2NH3(g)
can you please show the steps!
i think this is limiting reactants but i simply can't get to how to solve it!
the equation given is N2(g) + 3H2(g) -> 2NH3(g)
can you please show the steps!
i think this is limiting reactants but i simply can't get to how to solve it!
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General method:
Convert your reactants into moles, using molecular masses from the periodic table.
Use your balanced chemical equation to find out which is the limiting reagent.
Convert from moles of limiting reagent into moles of product, again using the balanced chemical equation.
Convert from moles of product into mass of product, using the periodic table to find the molecular mass.
N2(g) + 3H2(g) -> 2NH3(g)
mol N2 = mass N2 * molecular mass N2
= (1.00 x 10^3g N2)(1mol N2/28.02g N2)
= 35.7 mol N2 available
mol H2 = mass H2 * molecular mass H2
= (5.00 x 10^2g)(1mol H2/2.02g H2)
= 248mol H2 available
Now, the balanced chemical equation tells us in what mole ratios they react to form the product; that is, we need 3 times as much H2 as we need N2. Multiply the number of moles we actually have of N2, by the mole ratio between H2 and N2 (make sure N2 is on the bottom), to find out the minimal amount of moles of H2 we need to perform the reaction
(35.7mol N2)(3mol H2 / 1mol N2)
= 107mol H2 required
We can now see that we have quite a bit more H2 than we require to perform the reaction; H2 is our excess reagent. This means that the only other reagent, N2, must be the limiting reagent: once it runs out, no more product will be produced. Therefore, again using the mole ratios given by the chemical equation, we can convert the number of moles of our limiting reagent into a number of moles of product.
(35.7mol N2)(2mol NH3 / 1mol N2)
= 71.4mol NH3
Now, just multiply by the molecular mass of NH3 to find out how much that is in grams.
(71.4mol NH3)(17.04g NH3 / 1mol NH3)
= 1217g NH3
Key idea to remember: the quantities of reagents and products are related through the mole ratios given by the balanced chemical equation of the reaction.
Convert your reactants into moles, using molecular masses from the periodic table.
Use your balanced chemical equation to find out which is the limiting reagent.
Convert from moles of limiting reagent into moles of product, again using the balanced chemical equation.
Convert from moles of product into mass of product, using the periodic table to find the molecular mass.
N2(g) + 3H2(g) -> 2NH3(g)
mol N2 = mass N2 * molecular mass N2
= (1.00 x 10^3g N2)(1mol N2/28.02g N2)
= 35.7 mol N2 available
mol H2 = mass H2 * molecular mass H2
= (5.00 x 10^2g)(1mol H2/2.02g H2)
= 248mol H2 available
Now, the balanced chemical equation tells us in what mole ratios they react to form the product; that is, we need 3 times as much H2 as we need N2. Multiply the number of moles we actually have of N2, by the mole ratio between H2 and N2 (make sure N2 is on the bottom), to find out the minimal amount of moles of H2 we need to perform the reaction
(35.7mol N2)(3mol H2 / 1mol N2)
= 107mol H2 required
We can now see that we have quite a bit more H2 than we require to perform the reaction; H2 is our excess reagent. This means that the only other reagent, N2, must be the limiting reagent: once it runs out, no more product will be produced. Therefore, again using the mole ratios given by the chemical equation, we can convert the number of moles of our limiting reagent into a number of moles of product.
(35.7mol N2)(2mol NH3 / 1mol N2)
= 71.4mol NH3
Now, just multiply by the molecular mass of NH3 to find out how much that is in grams.
(71.4mol NH3)(17.04g NH3 / 1mol NH3)
= 1217g NH3
Key idea to remember: the quantities of reagents and products are related through the mole ratios given by the balanced chemical equation of the reaction.