1.Given that 1g of hydrogen combines completely with 8g of oxygen to form water, how much water can you make with 9g of hydrogen and 32g of oxygen?
We want to find which of these two elements is our limiting amount.
Here's some helpful equations to consider.
8x = 32 //The equation for oxygen.
1x = 9 //The equation for hydrogen.
From these two values, we want to determine the value of x that makes each equation true.
For the first equation, x = 4 and for the second, x = 9.
Because 4 is lower, this means oxygen is our limiting reagent and we can only make 4 molecules of water with 5g of hydrogen left over.
To calculate how much water can be made with 9g of hydrogen and 32g of oxygen, we need to determine the limiting reactant.
1. Start by calculating the molar mass of each element:
- Hydrogen (H2): 2 x 1.01 g/mol = 2.02 g/mol
- Oxygen (O2): 2 x 16.00 g/mol = 32.00 g/mol
2. Next, calculate the moles of each element given their masses:
- Moles of hydrogen: 9g / 2.02 g/mol = 4.455 moles
- Moles of oxygen: 32g / 32.00 g/mol = 1 mole
3. Now, determine the mole ratio between hydrogen and water. From the given information, we know that 1 gram of hydrogen combines completely with 8 grams of oxygen to form water. This means that 2 moles of hydrogen combine with 1 mole of oxygen to form 2 moles of water.
4. We can calculate the maximum moles of water that can be formed by comparing the moles of hydrogen and oxygen available. In this case, we have 4.455 moles of hydrogen and 1 mole of oxygen. Since the mole ratio is 2:1, we take the limiting reactant, which is oxygen.
5. Finally, calculate the maximum moles of water formed using the limiting reactant:
- Moles of water: 1 mole (limiting reactant) x 2 moles of water/1 mole of oxygen = 2 moles of water
Based on the calculations, with 9g of hydrogen and 32g of oxygen, the maximum amount of water that can be formed is 2 moles. To convert this into grams, we need to multiply the moles of water by the molar mass of water:
Molar mass of water (H2O): 2 x 1.01 g/mol (Hydrogen) + 16.00 g/mol (Oxygen) = 18.02 g/mol
So, 2 moles of water x 18.02 g/mol = 36.04 grams of water.
Therefore, with 9g of hydrogen and 32g of oxygen, you can make 36.04 grams of water.
To find out how much water can be made with 9g of hydrogen (H2) and 32g of oxygen (O2), we need to determine the limiting reactant of the reaction. The limiting reactant is the one that gets completely consumed and determines the maximum amount of product that can be formed.
First, let's calculate the number of moles of hydrogen and oxygen using their molar masses. The molar mass of hydrogen (H2) is 2 g/mol, and the molar mass of oxygen (O2) is 32 g/mol.
Number of moles of hydrogen = mass of hydrogen / molar mass of hydrogen
Number of moles of hydrogen = 9g / 2g/mol
Number of moles of hydrogen = 4.5 mol
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
Number of moles of oxygen = 32g / 32g/mol
Number of moles of oxygen = 1 mol
Now, we can use the balanced chemical equation to determine the stoichiometry of the reaction. The balanced equation for the formation of water (H2O) from hydrogen and oxygen is:
2H2 + O2 -> 2H2O
According to the equation, 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. Therefore, the ratio of hydrogen to oxygen is 2:1.
Since we have 4.5 moles of hydrogen and 1 mole of oxygen, we find that the oxygen is the limiting reactant because we need only 1 mole of oxygen to react with 2 moles of hydrogen.
Now, let's calculate the maximum amount of water that can be formed using the limiting reactant, which is oxygen.
Number of moles of water = (moles of limiting reactant) × (moles of water per mole of oxygen)
Number of moles of water = 1 mol × 2 mol of water per 1 mol of oxygen
Number of moles of water = 2 mol
Finally, let's calculate the mass of water formed using the molar mass of water, which is 18 g/mol.
Mass of water = (number of moles of water) × (molar mass of water)
Mass of water = 2 mol × 18 g/mol
Mass of water = 36 g
Therefore, with 9g of hydrogen and 32g of oxygen, you can make a maximum of 36g of water.