Please help..
1. How much water could you get if you started with 250.0 grams of hydrogen?
2. How much water could you get if you started with 250.0 grams of oxygen?
3. How much propane would you need to produce 1000.0 grams of water?
4. How much of each reactant is needed to produce 150 grams of hydrogen gas?
2H2 + O2 ---> 2H2O
You get 1 mol water for each 1 mol of H2
H2O is 2+16 = 18 grams / mol
H2 is 2 grams / mol so 250 grams = 125 mol
125 mol * 18 grams/mol = 2250 grams of H2O
Now you try.
1. To determine how much water can be obtained from 250.0 grams of hydrogen, we need to consider the balanced chemical equation for the reaction of hydrogen and oxygen to form water, which is 2H2 + O2 -> 2H2O. From the equation, we can see that for every two moles of hydrogen, we get two moles of water. This means that 1 mole of hydrogen corresponds to 1 mole of water.
The molar mass of hydrogen is approximately 1 gram/mole, so 250.0 grams of hydrogen is equal to 250.0 moles of hydrogen. Since each mole of hydrogen corresponds to 1 mole of water, we can conclude that 250.0 grams of hydrogen would produce 250.0 grams of water.
2. Similarly, to determine how much water can be obtained from 250.0 grams of oxygen, we need to consider the balanced chemical equation. Using the same equation as in question 1 (2H2 + O2 -> 2H2O), we can see that for every one mole of oxygen, we get two moles of water.
The molar mass of oxygen is approximately 16 grams/mole, so 250.0 grams of oxygen is equal to 15.625 moles of oxygen. Since each mole of oxygen corresponds to two moles of water, we can conclude that 250.0 grams of oxygen would produce 31.25 grams of water.
3. To determine how much propane would be needed to produce 1000.0 grams of water, we need to consider the balanced chemical equation for the combustion of propane. The balanced equation is C3H8 + 5O2 -> 3CO2 + 4H2O.
From the equation, we can see that for every 4 moles of water, we need 1 mole of propane. Therefore, in order to determine the amount of propane needed, we first need to convert grams of water to moles of water.
The molar mass of water is approximately 18 grams/mole, so 1000.0 grams of water is equal to 55.56 moles of water. Since the mole ratio of water to propane is 4:1, we divide the number of moles of water by 4:
55.56 moles water / 4 = 13.89 moles propane
So, you would need approximately 13.89 moles of propane to produce 1000.0 grams of water.
4. To determine how much of each reactant is needed to produce 150 grams of hydrogen gas, we need to consider the balanced chemical equation for the reaction of hydrogen and oxygen to form water. The balanced equation is:
2H2 + O2 -> 2H2O
From the equation, we can see that for every two moles of hydrogen gas, we need one mole of oxygen gas to produce two moles of water.
The molar mass of hydrogen is approximately 2 grams/mole, so 150 grams of hydrogen is equal to 75 moles of hydrogen. Since the mole ratio of hydrogen to oxygen is 2:1, we divide the number of moles of hydrogen by 2:
75 moles H2 / 2 = 37.5 moles O2
So, you would need approximately 37.5 moles of oxygen gas and 75 moles of hydrogen gas to produce 150 grams of hydrogen gas.
Sure, I can help you with these questions. To get the answers, we need to use the stoichiometry of the chemical reactions involved. Let's go through each question one by one.
1. To determine how much water you could get from 250.0 grams of hydrogen, we need to know the balanced chemical equation for the reaction between hydrogen and oxygen to form water. The balanced equation is:
2H₂ + O₂ -> 2H₂O
From the equation, we see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. To calculate the amount of water produced, we need to convert grams of hydrogen to moles using the molar mass of hydrogen (1.008 g/mol).
First, calculate the number of moles of hydrogen:
250.0 g / 1.008 g/mol = 248.0 mol
Since 2 moles of hydrogen react to form 2 moles of water, we can convert moles of hydrogen to moles of water:
248.0 mol H₂ = 248.0 mol H₂O
Finally, convert moles of water to grams:
248.0 mol H₂O x (18.015 g/mol) = 4464.12 g
Therefore, you could get approximately 4464.12 grams of water from 250.0 grams of hydrogen.
2. To determine how much water you could get from 250.0 grams of oxygen, we can follow a similar approach as in question 1, using the balanced chemical equation:
2H₂ + O₂ -> 2H₂O
From the equation, we see that 1 mole of oxygen reacts with 2 moles of hydrogen to produce 2 moles of water. Start by converting grams of oxygen to moles using the molar mass of oxygen (16.00 g/mol).
Calculate the number of moles of oxygen:
250.0 g / 16.00 g/mol = 15.625 mol
Since 1 mole of oxygen reacts to form 2 moles of water, we can convert moles of oxygen to moles of water:
15.625 mol O₂ x (2 mol H₂O / 1 mol O₂) = 31.25 mol H₂O
Finally, convert moles of water to grams:
31.25 mol H₂O x (18.015 g/mol) = 562.93 g
Therefore, you could get approximately 562.93 grams of water from 250.0 grams of oxygen.
3. To determine how much propane is needed to produce 1000.0 grams of water, we need to know the balanced chemical equation for the reaction between propane (C₃H₈) and oxygen (O₂) to form water. The balanced equation is:
C₃H₈ + 5O₂ -> 4H₂O + 3CO₂
From the equation, we see that 1 mole of propane reacts with 5 moles of oxygen to produce 4 moles of water. Start by converting grams of water to moles using the molar mass of water (18.015 g/mol).
Calculate the number of moles of water:
1000.0 g / 18.015 g/mol = 55.51 mol
Since 1 mole of propane reacts to form 4 moles of water, we can convert moles of water to moles of propane:
55.51 mol H₂O x (1 mol C₃H₈ / 4 mol H₂O) = 13.88 mol C₃H₈
Finally, convert moles of propane to grams using the molar mass of propane (44.097 g/mol):
13.88 mol C₃H₈ x (44.097 g/mol) = 612.90 g
Therefore, you would need approximately 612.90 grams of propane to produce 1000.0 grams of water.
4. To determine how much of each reactant is needed to produce 150 grams of hydrogen gas, we need to know the balanced chemical equation for the reaction of the reactants that produce hydrogen gas. Without the specific reaction, it is not possible to provide an accurate answer. However, I can guide you through the process once you provide the balanced chemical equation for the reaction.