What is the largest amount of H2O in grams that can be formed when 7.30 g of H2 reacts with 7.30 g of O2, based on the balanced reaction below.
2 H2 + O2 2 H2O
first find the limiting agent because this determines the amount of water produced. this is done by calculating the mole of each gas and see which has a mole that satisfy the reaction;
for hydrogen; 7.30/2 = 3.65moles
for oxygen; 7.30/32 = 0.228moles
so hydrogen is excess in amount; the required mole is 2 (according to the equation).
oxygen is less than expected so it is the limiting agent which determines the amount of water produced.
use the mole for oxygen to determine the mole of water. then find the mass by multiplying the molar mass of water with its mole.
hope that helps.
To find the largest amount of H2O that can be formed, we need to use stoichiometry and the given masses of H2 and O2. Here are the steps to calculate it:
1. Calculate the number of moles of H2 and O2 using their respective molar masses.
- Molar mass of H2 = 2.02 g/mol
- Molar mass of O2 = 32.00 g/mol
Moles of H2 = mass of H2 / molar mass of H2
= 7.30 g / 2.02 g/mol
= 3.6139 mol
Moles of O2 = mass of O2 / molar mass of O2
= 7.30 g / 32.00 g/mol
= 0.2281 mol
2. Determine the limiting reagent (the reactant that is completely consumed first) by comparing the moles of H2 and O2. The reactant with the smaller number of moles is the limiting reagent.
In this case, O2 has fewer moles (0.2281 mol) compared to H2 (3.6139 mol). Therefore, O2 is the limiting reagent.
3. Use stoichiometry to calculate the moles of H2O formed.
From the balanced equation, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.
Moles of H2O = (moles of O2) * (2 moles of H2O / 1 mole of O2)
= 0.2281 mol * (2 mol/ 1 mol)
= 0.4562 mol
4. Convert the moles of H2O to grams by multiplying it by the molar mass of H2O.
Molar mass of H2O = 18.02 g/mol
Mass of H2O = moles of H2O * molar mass of H2O
= 0.4562 mol * 18.02 g/mol
= 8.229 g
Therefore, the largest amount of H2O that can be formed is 8.229 grams.
To determine the largest amount of H2O that can be formed, we need to use stoichiometry, which involves using the balanced equation to relate the moles of reactants and products.
Given the balanced equation:
2 H2 + O2 -> 2 H2O
First, we need to convert the given masses of H2 and O2 into moles.
1. Calculate the number of moles of H2:
Using the molar mass of hydrogen (H2), which is 2.02 g/mol, we can calculate the moles of H2:
7.30 g H2 × (1 mol H2 / 2.02 g H2) = 3.62 mol H2
2. Calculate the number of moles of O2:
Using the molar mass of oxygen (O2), which is 32.00 g/mol, we can calculate the moles of O2:
7.30 g O2 × (1 mol O2 / 32.00 g O2) = 0.23 mol O2
Next, we examine the stoichiometric ratio between H2 and H2O to determine the limiting reactant. The limiting reactant is the one that is completely consumed and restricts the amount of product formed.
From the balanced equation, we see that 2 moles of H2 forms 2 moles of H2O. Therefore, the stoichiometric ratio between H2 and H2O is 2:2, which simplifies to 1:1.
To determine the limiting reactant, we compare the moles of H2 and O2. In this case, since the moles of H2 are larger, H2 is in excess, and O2 is the limiting reactant.
Based on the stoichiometric ratio, we can determine the moles of H2O that can be formed:
Since 1 mol O2 reacts to form 2 mol H2O, and we have 0.23 mol O2 (the limiting reactant), we can calculate the moles of H2O:
0.23 mol O2 × (2 mol H2O / 1 mol O2) = 0.46 mol H2O
Finally, we need to convert the moles of H2O to grams.
Using the molar mass of H2O, which is 18.02 g/mol, we can calculate the grams of H2O:
0.46 mol H2O × (18.02 g H2O / 1 mol H2O) = 8.29 g H2O
Therefore, the largest amount of H2O that can be formed when 7.30 g of H2 reacts with 7.30 g of O2 is 8.29 grams.