Given the following:
4 KO2 + 2 H2O ===> 4 KOH + 3 O2
Find the grams KO2 and H2O needed to make 1350 grams O2
Given, The equilibrium reaction,
4KO2 + 2H2O 4KOH + 3O2
Also given,
Mass of O2 = 1350 grams
Now, calculating the number of moles of O2,
= 1350 g of O2 x ( 1 mol /31.998 g)
= 42.1901 mol O2
Now, Using the moles of O2 and the mole ratio from the balanced chemical reaction, calculating the number of moles KO2 and
= 42.1901 mol O2 x ( 4 mol KO2 / 3 mol O2)
= 56.2535 mol KO2
Converting the number of moles to grams,
= 56.2535 mol KO2 x (71.1 g / 1 mol)
= 3999.6 g of KO2 Or 4.00 x 103 g of KO2 [ 3 S.F]
Similarly,
= 42.1901 mol O2 x ( 2 H2O / 3 mol O2)
= 28.1268 mol H2O
Converting the number of moles to grams,
= 28.1268 mol H2O x (18.02 g / 1 mol)
= 506.8 g of H2O Or 507 g of H2O [ 3S.F]
Question 2 Consider the following reaction: 4 KO2(s) + 2 H2O(1) ® 4 KOH(s) + 3 O2(g) Suppose 83 grams of KO2 are added to 27 grams of H20. What is the limiting reactant and how many grams of KOH is produced (theoretically)? Answer below and show your work on the work sheet. Label your work as #2. HTML
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