a mixture of .0285g of H and .0141 mol of oxygen in a closed container is sparked to initiate a reaction. How many grams of H2O can form?Which reactant is in excess and how many grams remain?
Write and balance the equation.
Convert g Hydrogen to moles. moles = grams/molar mass
You have oxygen in moles.
Then take one element at a time and determine the moles H2O formed. Follow the instructions on this worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Basically, you take H (and ignore oxygen) and calculate the moles H2O formed.
Then take oxygen (and ignore hydrogen) and calculate moles H2O formed.
You will obtain two different answers and only one can be right; the correct one in limiting reagent problems is ALWAYS the smaller value. Then take that value of moles and convert mole to grams. grams = moles x molar mass.
Post your work if you get stuck.
To determine how many grams of H2O can form and which reactant is in excess, we need to follow these steps:
1. Write the balanced chemical equation for the reaction between hydrogen (H2) and oxygen (O2) to form water (H2O).
2 H2 + O2 → 2 H2O
2. Calculate the number of moles for each reactant:
- Moles of H (H2): Given that the mass of H is 0.0285g, the molar mass of H is 1 g/mol. Hence, the number of moles of H is:
moles of H = mass of H / molar mass of H
moles of H = 0.0285g / 1 g/mol
moles of H = 0.0285 mol
- Moles of O (O2): Given that the number of moles of O2 is 0.0141 mol, we don't need to perform any conversion.
3. Using the balanced chemical equation, determine the mole ratio between H2 and O2. In this case, the ratio is 2:1 since 2 moles of H2 react with 1 mole of O2.
4. Compare the moles of H2 and O2. If the moles of either reactant are higher, that reactant is in excess. If the moles of H2 are less than half the moles of O2, it means there is an excess of O2. If the moles of H2 are more than double the moles of O2, it means there is an excess of H2.
5. Calculate the limiting reactant to determine how much H2O can form. Since we have a 2:1 mole ratio, the limiting reactant can be determined by comparing the moles of H2 to the moles of O2.
- Moles of H2O formed = Moles of limiting reactant * Mole ratio (according to the balanced equation)
6. Calculate the moles of H2O formed:
moles of H2O = moles of limiting reactant * mole ratio (from the balanced equation)
moles of H2O = 0.0141 mol * (2 mol H2O / 2 mol H2)
moles of H2O = 0.0141 mol
7. Convert moles of H2O to grams:
mass of H2O = moles of H2O * molar mass of H2O
mass of H2O = 0.0141 mol * 18 g/mol
mass of H2O = 0.2538 g
Therefore, 0.2538 grams of H2O can form.
8. Calculate the remaining excess reactant:
- If H2 is in excess:
moles of excess H2 = moles of H2 - (moles of H2O formed * mole ratio of H2)
mass of excess H2 = moles of excess H2 * molar mass of H2
- If O2 is in excess:
moles of excess O2 = moles of O2 - (moles of H2O formed * mole ratio of O2)
mass of excess O2 = moles of excess O2 * molar mass of O2
After performing the calculations, you can determine which reactant is in excess and calculate the remaining mass in grams.