What is the largest amount of H2O in grams that can be formed when 5.70 g of H2 reacts with 5.70 g of O2, based on the balanced reaction below.
2 H2 + O2 2 H2O
To find the largest amount of H2O that can be formed, we need to determine the limiting reactant between H2 and O2. The limiting reactant is the one that is completely consumed in the reaction, thus determining the maximum amount of product that can be formed.
Let's begin by calculating the number of moles of H2 and O2 using their respective molar masses.
The molar mass of H2 is 2 g/mol, so we can calculate the number of moles of H2 using the given mass of 5.70 g:
moles of H2 = 5.70 g / 2 g/mol = 2.85 mol
The molar mass of O2 is 32 g/mol, so we can calculate the number of moles of O2 using the given mass of 5.70 g:
moles of O2 = 5.70 g / 32 g/mol = 0.17813 mol
Now, we can determine the limiting reactant by comparing the moles of each reactant to their stoichiometric coefficients in the balanced equation. According to the balanced equation, it requires 2 moles of H2 for every 1 mole of O2 to react completely:
H2 : O2 = 2.85 mol : 0.17813 mol
Dividing both ratios by the smallest ratio (0.17813 mol), we get:
H2 : O2 = 16 : 1
Since the stoichiometric ratio of H2 to O2 is 16:1, it means there is an excess of H2. Therefore, H2 is the limiting reactant, and O2 is in excess.
Now, let's calculate the maximum moles of H2O that can be formed based on the limiting reactant. From the balanced equation, we know that the stoichiometric ratio between H2 and H2O is 2:2 (or 1:1):
moles of H2O = moles of H2 = 2.85 mol
Finally, we can convert the moles of H2O to grams using the molar mass of H2O, which is 18 g/mol:
mass of H2O = moles of H2O x molar mass of H2O
mass of H2O = 2.85 mol * 18 g/mol = 51.3 g
Therefore, the largest amount of H2O that can be formed when 5.70 g of H2 reacts with 5.70 g of O2 is 51.3 grams.
To find the largest amount of H2O that can be formed when 5.70 g of H2 reacts with 5.70 g of O2, we need to determine the limiting reactant. The limiting reactant is the one that gets completely consumed and determines the maximum amount of product that can be formed.
1. First, let's calculate the number of moles for each reactant using their molar masses:
Moles of H2 = mass of H2 / molar mass of H2
= 5.70 g / 2.02 g/mol (molar mass of H2)
= 2.8218 mol
Moles of O2 = mass of O2 / molar mass of O2
= 5.70 g / 32.00 g/mol (molar mass of O2)
= 0.1781 mol
2. Next, let's determine the stoichiometry of the balanced equation.
From the balanced equation:
2 H2 + O2 -> 2 H2O
We can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.
3. Now, let's compare the ratio of moles actually available to the stoichiometric ratio to find the limiting reactant.
For H2, the ratio is:
Moles of H2O / Moles of H2 = 2 / 2 = 1
For O2, the ratio is:
Moles of H2O / Moles of O2 = 2 / 1 = 2
Since the ratio is higher for O2, it means that O2 is the limiting reactant.
4. Now, let's calculate the moles and mass of H2O formed using the limiting reactant, O2.
Moles of H2O = Moles of limiting reactant (O2) * Stoichiometric ratio (moles of H2O / moles of O2)
= 0.1781 mol * (2 mol H2O / 1 mol O2)
= 0.3562 mol
Mass of H2O = Moles of H2O * Molar mass of H2O
= 0.3562 mol * 18.02 g/mol (molar mass of H2O)
= 6.420 g
Therefore, the largest amount of H2O that can be formed is approximately 6.420 grams.
I don't see an arrow.
This is a limiting reagent problem; I know that because amounts for BOTH reactants are given.
Convert 5.70 g O2 and 5.70 g H2 to moles. moles = grams/molar mass.
First O2 and ignore H2.
Convert moles O2 to moles H2O using the coefficients in the balanced equation.
Next H2 and ignore O2.
Convert moles H2 to moles H2O using the coefficients in the balanced equation.
You have two answers for the same product; both can't be correct. The correct value, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Now convert the moles (the smaller value) to grams. g = moles x molar mass.
Here is a worked example of a simple stoichiometry problem. It will solve all of your simple stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html