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Starting with 195 g Li2O and 106 g H2O, decide which reactant is present in limiting quantities. Given: Li2O+H2O→2LiOH

User A Toll
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1 Answer

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Final answer:

To determine the limiting reagent in a chemical reaction, calculate the moles of each reactant and compare it to the stoichiometric ratio in the balanced equation. In this case, the limiting reagent is H2O due to the lesser amount of moles compared to Li2O.

Step-by-step explanation:

The limiting reagent in a chemical reaction is the reactant that is completely consumed, limiting the amount of product that can be formed. To determine the limiting reagent, you need to calculate the moles of each reactant and compare it to the stoichiometric ratio in the balanced equation. In this case, you have 195 g of Li2O and 106 g of H2O. Converting these masses to moles, you can then compare the ratio of moles to determine which reactant is present in limiting quantities.

First, calculate the moles of each reactant:

1 mol Li2O = 29.88 g Li2O

1 mol H2O = 18.02 g H2O

Using these conversion factors, we can calculate the moles of Li2O and H2O:

Li2O: 195 g Li2O × (1 mol Li2O/29.88 g Li2O) = 6.52 mol Li2O

H2O: 106 g H2O × (1 mol H2O/18.02 g H2O) = 5.88 mol H2O

Next, we compare the molar ratio of Li2O to H2O in the balanced equation, which is 1:1. Since the moles of Li2O (6.52 mol) is greater than the moles of H2O (5.88 mol), the limiting reagent is H2O.

User Vlasits
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