identify limiting reagent in given reaction 2H2O + O2 = 2H2O
1) 1mol H2 and 1mol O2 are made to react
2) 10mol H2 and 5mol O2 are made to react
3) 5.6L of H2 and 11.2L of O2 are made to react
4) 10g of H2 and 64g of O2 are made to react
1) 1mol H2 and 1mol O2 are made to react
2) 10mol H2 and 5mol O2 are made to react
3) 5.6L of H2 and 11.2L of O2 are made to react
4) 10g of H2 and 64g of O2 are made to react
Answers & Comments
Verified answer
Explanation:
The balanced chemical equation for the reaction 2H2O + O2 = 2H2O shows that the ratio of H2O to O2 required for the reaction is 2:1. This means that whichever reactant is present in a ratio other than 2:1 with the other will be the limiting reagent.
1) For 1 mol H2 and 1 mol O2, the ratio of H2O to O2 is 2:1. Therefore, neither H2 nor O2 is in excess, and neither is the limiting reagent.
2) For 10 mol H2 and 5 mol O2, the ratio of H2O to O2 is 2:1. Therefore, O2 is present in a ratio of 5:1 with respect to H2O and is the limiting reagent.
3) For 5.6 L of H2 and 11.2 L of O2, we need to convert the volumes to moles to determine the limiting reagent. Assuming standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 L. Therefore, we have:
- 5.6 L of H2 = 5.6/22.4 = 0.25 mol H2
- 11.2 L of O2 = 11.2/22.4 = 0.5 mol O2
The ratio of H2O to O2 required for the reaction is 2:1. Therefore, we need 0.25 mol H2O and 0.125 mol O2 for the reaction. Since O2 is present in excess (0.5 mol > 0.125 mol), H2 is the limiting reagent.
4) For 10 g of H2 and 64 g of O2, we need to convert the masses to moles to determine the limiting reagent. Using the molar mass of H2 (2 g/mol) and O2 (32 g/mol), we have:
- 10 g of H2 = 10/2 = 5 mol H2
- 64 g of O2 = 64/32 = 2 mol O2
The ratio of H2O to O2 required for the reaction is 2:1. Therefore, we need 2.5 mol H2O and 1.25 mol O2 for the reaction. Since O2 is present in excess (2 mol > 1.25 mol), H2 is the limiting reagent.
Answer:
The ratio of H2O to O2 required for the reaction is 2:1. Therefore, we need 2.5 mol H2O and 1.25 mol O2 for the reaction. Since O2 is present in excess (2 mol > 1.25 mol), H2 is the limiting reagent.