Question

8gram of hydrogen and oxygen to from water according to following equation to find limiting reactant ?

Answer 1

Answer:

The reaction of hydrogen gas and oxygen gas to make water is as follows:

2H2(g)+O2(g)=2H2O(l)2H2(g)+O2(g)=2H2O(l)

8g of O2 gas is about 8/32g/mol=.25molesofO2(g)8/32g/mol=.25molesofO2(g)

8g of H2 gas is about 8/2g/mol=4molesofH2(g)8/2g/mol=4molesofH2(g)

We can calculate the amount of product (water) that each reactant will make, given our moles above, and whichever makes lesser moles of product is our limiting reactant.

.25molesO2∗(2molesH2O/1moleO2)=.5molesofH2O(l).25molesO2∗(2molesH2O/1moleO2)=.5molesofH2O(l)

4molesH2∗(2molesH2O/2molesH2)=4molesofH2O(l)4molesH2∗(2molesH2O/2molesH2)=4molesofH2O(l)

As you can see above, oxygen gas is the limiting reactant. Note: none of these numbers are to proper amounts of significant figures.

Explanation:

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