1.5g of H reacts with14.5g Oxygen
1)which is the limiting reagent?
2)calculate the amount of reactant which remains unreacted?
the product is water (H2O)
the reaction ratio is by moles ... 2 moles H to 1 mole O
convert the grams to moles (divide by atomic mass)
To determine the limiting reagent, we need to compare the moles of each reactant to their stoichiometric coefficients.
1) First, let's calculate the number of moles for each reactant:
Molar mass of H (hydrogen) = 1g/mol
Molar mass of O2 (oxygen) = 2 * 16g/mol = 32g/mol
Moles of H = mass of H / molar mass of H
= 1.5g / 1g/mol
= 1.5 mol
Moles of O2 = mass of O2 / molar mass of O2
= 14.5g / 32g/mol
= 0.453125 mol (rounded to 6 decimal places)
2) Next, we compare the mole ratio of H to O2 in the balanced equation to see which reactant is limiting:
The balanced equation for the reaction between H and O2 is:
2 H2 + O2 → 2 H2O
According to the balanced equation, the mole ratio between H and O2 is 2:1.
Since the mole ratio between H and O2 is 2:1, we need twice as many moles of H2 as we have moles of O2 for a complete reaction.
The moles of O2 available (0.453125 mol) is less than half of the moles of H2 (1.5 mol). Therefore, O2 is the limiting reagent.
3) To calculate the amount of reactant that remains unreacted, we need to find the excess of the non-limiting reactant.
Since O2 is the limiting reagent, we will use its moles to calculate the amount of H2 that reacts.
Moles of H2 used = 0.5 * moles of O2 (from the 2:1 ratio)
Moles of H2 used = 0.5 * 0.453125 mol
= 0.2265625 mol (rounded to 7 decimal places)
To find the remaining moles of H2, subtract the moles used from the total moles of H2:
Remaining moles of H2 = Total moles of H2 - Moles of H2 used
= 1.5 mol - 0.2265625 mol
= 1.2734375 mol (rounded to 7 decimal places)
Finally, we can calculate the mass of H2 that remains unreacted:
Mass of H2 remaining = Remaining moles of H2 * molar mass of H2
= 1.2734375 mol * 2g/mol
= 2.547875g (rounded to 6 decimal places)
Therefore, the amount of H2 that remains unreacted is approximately 2.547875 grams.
To determine the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratio of the balanced chemical equation.
The balanced chemical equation for the reaction between hydrogen (H2) and oxygen (O2) is:
2H2 + O2 -> 2H2O
1) Find the moles of each reactant:
- Hydrogen (H2): 1.5g / 2g/mol = 0.75 moles
- Oxygen (O2): 14.5g / 32g/mol = 0.453 moles
2) Calculate the stoichiometric ratio (moles) between hydrogen and oxygen:
According to the balanced equation, it takes 2 moles of hydrogen to react with 1 mole of oxygen.
3) Compare the moles of each reactant to the stoichiometric ratio:
- Hydrogen: 0.75 moles
- Oxygen: 0.453 moles * (2 moles of hydrogen / 1 mole of oxygen) = 0.906 moles
From the comparison, we can see that the number of moles of oxygen is lower than the stoichiometric ratio. Therefore, oxygen is the limiting reagent.
To calculate the amount of reactant remaining unreacted, we need to determine the excess reactant.
4) Calculate the moles of the excess reactant:
- Hydrogen (H2): 0.75 moles - (0.453 moles * (2 moles of hydrogen / 1 mole of oxygen)) = 0.75 - 0.906 = -0.156 moles (negative value indicates the limiting reagent)
Since the moles of hydrogen are negative, it means that all of the hydrogen has reacted. Therefore, there is no hydrogen remaining as unreacted.
2H2 + O2 ==> 2H2O
mols H2 = gram/molar mass = 1.5/2 = 0.75
mols O2 = 14.5/32 = approx 0.45 but you need to do these values again and more accurately. All of my calculations are estimates.Convert H2 to mols H2O produced IF you had all of the oxygen needed (an excess of oxygen). That is 0.75 mols H2 x (2 mols H2O/2 mols H2) = 0.75 mols H2O produced.
Now convert 0.45 mols O2 to mols H2O produced IF you had all of the H2 needed (an excess of H2). That is 0.45 x (2 mols H2O/1 mol O2) = 0.9 mols H2O produced.
Compare the two values. In limiting reagent (LR) problems the smaller number is always the correct one; therefore, the reaction will produce 0.75 mols H2O and H2 is the LR. O2 is the excess reagent (ER). How much O2 is used in the reaction. That is 0.75 mols H2 x (1 mol O2/2 mols H2) = about 0.375 mols O2. So how much is left? That is 0.45 initially - 0.375 mols used = ? mols O2 remaining un-reacted. Post your work if you get stuck.