If 10.0 g of hydrogen gar reacts with 20.0 g of oxygen gas, how much water will be made?
So far this is all i have:
(10.0g H2/1)(2molH2/2.02H2) = 9.0099 mol H2
(20.0g O2/1)(2mol O2/32g O2) = 1.25 mol O2
With the limited reactant being O2
Now what do i do?
I also have to find the yield of only 20.0g of H20 that was made.
Recheck your math onthe moles of H2.
Yes, the limiting reactant was O2, it has less moles.
H2 + 1/2 02>>H2O
so, you only have 1.25moles O2, so you can use up to 2.50 moles H2, and you will get 2.5 moles water.
To determine the amount of water that will be made, you need to find the limiting reactant and use the stoichiometry of the balanced equation.
The balanced equation for the reaction between hydrogen gas (H2) and oxygen gas (O2) is:
2H2 + O2 → 2H2O
You have already determined that the number of moles of hydrogen gas (H2) is 9.0099 mol and the number of moles of oxygen gas (O2) is 1.25 mol.
To find the limiting reactant, you need to compare the stoichiometric ratios of the reactants. From the balanced equation, you can see that it requires 2 moles of H2 to react with 1 mole of O2 to produce 2 moles of H2O.
Now, calculate the moles of water that will be produced from each reactant:
From 9.0099 mol H2:
(9.0099 mol H2) x (2 mol H2O / 2 mol H2) = 9.0099 mol H2O
From 1.25 mol O2:
(1.25 mol O2) x (2 mol H2O / 1 mol O2) = 2.5 mol H2O
Since the stoichiometric ratio between H2 and O2 is 2:1, it means that for every 2 moles of H2, you need 1 mole of O2. However, you have more moles of H2 than necessary and less moles of O2. Therefore, the limiting reactant is O2.
To find the yield of water from only 20.0 g of H2O, you need to calculate the number of moles of water produced from 20.0 g of H2O:
(20.0 g H2O) / (18.015 g/mol H2O) ≈ 1.1108 mol H2O
Since the moles of water produced from O2 is the limiting reactant, the yield of water will be 2.5 mol H2O. Since 1.1108 mol H2O is less than 2.5 mol H2O, the actual yield will be 1.1108 mol H2O.
To determine the amount of water produced, you need to find the balanced chemical equation for the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O). The balanced equation is:
2H2 + O2 -> 2H2O
From the balanced equation, you can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.
To calculate the amount of water produced, you need to compare the number of moles of H2 and O2 and determine the limiting reactant.
In your calculations, you found that you have 9.0099 mol of H2 and 1.25 mol of O2.
Since the stoichiometric ratio between H2 and O2 is 2:1 (2 moles of H2 react with 1 mole of O2), you need half the number of moles of O2 compared to H2. In this case, 9.0099 moles of H2 would require 4.5049 moles of O2 to react fully.
However, you only have 1.25 moles of O2 available, which is less than what is required to react with all the H2. Therefore, O2 is the limiting reactant.
To find the number of moles of H2O produced, you can use the stoichiometric ratio between O2 and H2O, which is also 2:1. So for every 1 mole of O2, you get 2 moles of H2O.
Since you have 1.25 moles of O2, you can calculate the moles of H2O produced:
1.25 mol O2 * (2 mol H2O / 1 mol O2) = 2.5 mol H2O
Therefore, 2.5 moles of water will be produced when 10.0 g of hydrogen gas reacts with 20.0 g of oxygen gas.
If you want to find the yield of only 20.0 g of water that was produced, you can use the molar mass of water to convert the grams of water to moles.
The molar mass of water (H2O) is approximately 18.015 g/mol.
To convert grams of water to moles, you can use the formula:
moles = mass / molar mass
moles = 20.0 g / 18.015 g/mol ≈ 1.110 moles
Therefore, the yield is approximately 1.110 moles of water when 20.0 g of water is produced.