what mass of water can be obtained from 4.0g of H2 and 16g of O2?
Find the equation:
H2 + O2 --> H2O
Balance the equation:
2H2 + O2 --> 2H2O
Find out how many moles of H2 and O2 you've got:
4.0g H2 * 1mol H2/2g = 2 mol H2
16g O2 * 1mol O2/32g = .5mol O2
Which one will be used up before the other (aka which one is the limiting reactant?)? You can use stoichiometry to find out:
2 mol H2 = 2 mol H2/1 mol O2
1 mol of O2 is required to use up all the H2 you have. You only have .5 mol of O2. Therefore, O2 is the limiting reactant. Now you can use stoichiometry to find how much water will be produced:
.5mol O2 = 1mol O2/2mol H2O
1 mol of H2O will be produced. Now convert it to grams:
1 mol H2O * 18g/mol = 18g H2O
This could have been reasoned through without all that extra work, but that's a general method to use when the numbers become less simple. Also, I wonder if there's any better methods of formatting as I'm sure that looks confusing.
H2 ?
To determine the mass of water that can be obtained from 4.0g of H2 and 16g of O2, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between hydrogen and oxygen to form water.
The balanced chemical equation for the reaction is:
2H2 + O2 → 2H2O
From the equation, we can see that 2 moles of hydrogen (H2) react with 1 mole of oxygen (O2) to produce 2 moles of water (H2O).
Step 1: Convert the given masses of H2 and O2 to moles.
Molar mass of H2 = 2g/mol
Number of moles of H2 = mass(m) / molar mass(M) = 4.0g / 2g/mol = 2.0 mol
Molar mass of O2 = 32g/mol
Number of moles of O2 = mass(m) / molar mass(M) = 16g / 32g/mol = 0.5 mol
Step 2: Determine the limiting reactant.
To find out which reactant is limiting, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation.
For every 2 moles of H2, we need 1 mole of O2.
Since we have 2.0 moles of H2 and 0.5 moles of O2, we can see that O2 is the limiting reactant because it is present in a lower amount compared to the stoichiometric ratio.
Step 3: Calculate the moles of water produced.
From the balanced equation, we know that 1 mole of O2 reacts to produce 2 moles of H2O.
So, for 0.5 moles of O2:
Number of moles of H2O = (0.5 mol O2) x (2 mol H2O / 1 mol O2) = 1.0 mol H2O
Step 4: Convert moles of water to grams.
Molar mass of H2O = 18g/mol
Mass of H2O = moles x molar mass = 1.0 mol x 18g/mol = 18g
Therefore, from 4.0g of H2 and 16g of O2, you can obtain 18g of water.
2H2+O2->2H2O
-convert g of H2 and of O2 each to mol H2O
steps 2 & 3:
http://www.jiskha.com/science/chemistry/stoichiometry.html
-convert the one with lower mol H2O to grams
-the reactant that produces the lesser amount of H2O is the limiting reagent and is the mass of water can be obtained
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm