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3g of H2 react with 29g of O2 to yield H2O.which is the limiting reactant? Calculate the maximum amount of H2O that can be formed. Calculate the amount the reactant which remains unreacted.
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The amount of the reactant which remains unreacted is 5g. Given that 3 g of hydrogen is the limiting reagent.
2H2(g) + O2 → H2O
In the equation, we can conclude that 2 mole H2 reacts with 1 mole O2
We know that,
- Molar mass of H2 = 2g
- Molar mass of O2 = 32g
Further,
4 g H2 reacts with 32 g O2
3 g H2 reacts with (32/4) x 3 g of O2 gas
= 24g
H2 is the limiting reagent and O2 is the excess reagent since the amount of O2 provided is greater than what is needed. 2 moles of hydrogen gas combine to generate 2 moles of water as a result.
4 g of H2 produces 36 g of water
Amount of H2O produced by 3 g H2 = (36/4) x 3 = 27g
Therefore, the reaction will result in the production of 27 g of water. Since 29 g of oxygen was provided and 24 g was used in the reaction, the remaining amount of oxygen gas is (29-24) = 5g
Summary:
3g of H2 react with 29g of O2 to yield H2O.which is the limiting reactant? Calculate the maximum amount of H2O that can be formed. Calculate the amount the reactant which remains unreacted.
3g of H2 react with 29g of O2 to yield H2O the limiting reactant is hydrogen. The maximum amount of water that can be formed is 27 g. The amount of the reactant which remains unreacted is 5g.
