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example of limiting reactant grams to mole

Sagot :

jafif67556

Answer:

Sure! Let's consider the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O).

The balanced chemical equation for this reaction is:

2H2 + O2 -> 2H2O

Let's say we have 10 grams of hydrogen gas (H2) and 20 grams of oxygen gas (O2). We can convert these masses to moles using the molar mass of each substance.

The molar mass of hydrogen gas (H2) is 2 grams/mol, and the molar mass of oxygen gas (O2) is 32 grams/mol.

To convert grams to moles, we use the formula:

moles = mass / molar mass

For hydrogen gas (H2):

moles of H2 = 10 grams / 2 grams/mol = 5 moles

For oxygen gas (O2):

moles of O2 = 20 grams / 32 grams/mol = 0.625 moles

Now, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.

From the balanced equation, we can see that the ratio of H2 to O2 is 2:1. This means that for every 2 moles of H2, we need 1 mole of O2.

Since we have 5 moles of H2 and 0.625 moles of O2, we can see that we have an excess of H2 and a deficiency of O2. Therefore, O2 is the limiting reactant.

To find the maximum amount of water (H2O) that can be formed, we use the mole ratio from the balanced equation. The ratio of H2O to O2 is 2:1.

moles of H2O = 0.625 moles of O2 x (2 moles of H2O / 1 mole of O2) = 1.25 moles of H2O

Finally, we can convert the moles of water (H2O) to grams using the molar mass of water, which is 18 grams/mol.

mass of H2O = moles of H2O x molar mass of H2O

mass of H2O = 1.25 moles x 18 grams/mol = 22.5 grams of H2O

Therefore, the maximum amount of water (H2O) that can be formed is 22.5 grams.

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