How do Find the Excess Reactant when one question will give me moles and another question will give me grams?

Answer 1

Since the coefficents of the substances in the balanced equation represent their number of moles, it is more practical to convert the amounts given in another units into mole to find the limiting and excess substances. Suppose the following question was given:

(1) If 5 moles of N2 gas reacts with 10 moles of H2 gas, how many grams of NH3 gas are produced? (H: 1 and N : 14)

Write the balanced equation:
N2(g) + 3H2(g) --------> 2NH3(g)

According to the equation 1 mole N2 reacts with 3 moles H2, then for 5 moles of N2, 15 moles of H2 are required. Since only 10 moles of H2 are available, H2 is limiting, N2 is excess. (the rest of the question is not solved, because your question involves only excess substance).

Now the same question was given in different form.

(2) If 5 moles of N2 gas reacts with 20 grams of H2 gas, how many grams of NH3 gas are produced? (H: 1 and N : 14)

N2(g) + 3H2(g) --------> 2NH3(g)
This time units are not the same. N2 : moles, H2 : grams. Covert the unit of H2 to mole by using its molar mass.
20 g / 2 g/mol = 10 moles.
Apply the procedure given in (1).

(3) If 140 grams of N2 gas reacts with 20 grams of H2 gas, how many grams of NH3 gas are produced? (H: 1 and N : 14)

N2(g) + 3H2(g) --------> 2NH3(g)

This time units are the same, but not mole. There are two ways: (a) Compare the moles, by converting the masses to moles.
N2 : 140 g / 28 g/mol = 5 moles
H2 : 20 g / 2 g/mol = 10 moles.
(b) Compare the masses, by converting the moles in the equation to grams. According to the equation;
1 mole N2(28 g) reacts with 3 moles of H2(3x2=6 g)
140 g N2 reacts with 140 x (6/28) = 30 g H2. Since only 20 g H2 are available H2 is limiting, N2 is excess.

(4) If 112 L of N2 gas at STP reacts with 20 grams of H2 gas, how many grams of NH3 gas are produced? (H: 1 and N : 14)

N2(g) + 3H2(g) --------> 2NH3(g)

Now, the volume of N2 was given. At STP 1 mole of any gas occupies 22.4 liters, Then,
Mole of N2 = 112 L / 22.4 L/mol = 5 mole.
Then compare the moles to find the limiting and excess reactants.

Answer 2

Al2S3 + 6H2O --> 2Al(OH)3 + 3H2S Having the balanced equation describing the reaction, the subsequent step is calculating the style of moles. The molecular weight of Al2S3 is what a million mole of its molecules weighs. this may well be the sum of the atomic weights of the atoms in the molecule. For Al2S3, the burden may well be 27*2 + 32*3 = one hundred fifty grams per mole. consequently 200 grams may well be 200 grams / (one hundred fifty grams/mole) = a million.333 moles. The style of moles of water available is one hundred fifty grams / (18 grams/mole) = 8.33 moles. Now, examine the molar ratios from the balanced equation. a million mole of Al2S3 demands six moles of H2O to form 3 moles of H2S. From the calculations above, a million.333 moles Al2S3 might require 8 moles of water. There are 8.33 moles available, so there is extra beneficial than sufficient to form the product. The molar ratio of Al2S3 to H2S is a million to 3, so a million.33 moles will form 4 moles of H2S. 4 moles of H2S weighs 4 moles * (34 grams/ mole) = 136 grams.

Answer 3

You must convert everything to moles in order to figure out which is the excess reactant based on the reaction. If you are already in moles, you miss the conversion step.

Answer 4

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