What is the largest amount of H2O in grams that can be formed when 6.10 g of H2 reacts with 6.10 g of O2, based on
2 H2 + O2 = 2 H2O
each mole of O2 requires 2 moles of H2.
How many moles of each reagent do you have?
For example, if you have 3 moles of H2 and 7 moles of O2, then the H2 will use 3/2 moles of O2, leaving extra O2 unused.
If you have 7 moles of H2 and 3 moles of O2, then all the O2 will be consumed by 6 moles of H2, leaving unused H2.
each mole of O2 produces 2 moles of H2O.
To find the maximum amount of H2O that can be formed when 6.10 g of H2 reacts with 6.10 g of O2, we need to determine the limiting reactant.
1. Start by determining the number of moles for each reactant:
H2 (hydrogen gas): 6.10 g ÷ molar mass of H2
O2 (oxygen gas): 6.10 g ÷ molar mass of O2
2. Next, compare the mole ratio in the balanced equation to determine which reactant is limiting. The balanced equation is:
2 H2 + O2 = 2 H2O
The ratio between H2 and O2 is 2:1. If the calculated moles of O2 are less than half of the calculated moles of H2, then O2 is the limiting reactant. Otherwise, H2 is the limiting reactant.
3. Once the limiting reactant is identified, we can use the moles of the limiting reactant to calculate the moles of H2O formed.
4. Finally, convert the moles of H2O to grams by multiplying them by the molar mass of H2O.
By following these steps, we can determine the maximum amount of H2O that can be formed.