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Chemistry 34 Online
OpenStudy (anonymous):

When heated, potassium chlorate decomposes in the following chemical reaction. potassium chlorate ==> potassium chloride + oxygen When 65.2 g of potassium chlorate is heated, 39.7 g of potassium chloride is produced. What mass of oxygen is produced?

OpenStudy (abmon98):

\[KCLO3-->KCL+O2 \] 1) Balance Chemical Equation 2)Use then Equation \[Number of Moles=Mass(g)/Molar Mass(g/mol)\] 3) Look for the Mole Ratio between the compounds found on the product side and on the reactant side Step 1 Write down the number of atoms that you have on each side of the equation \[KCLO3-->KCL+O2 \] On the reactant Side: 1 K, 1 Cl , 3 O On the product side: 1 K, 1 Cl, 2 O Add a Coeffcient of 3 to O2 and 2 to KCLO3 to Balance the oxygen together. \[2KCLO3-->KCL+3O2 \] On the reactant Side: 2 K, 2 Cl , 6 O On the product side: 1 K, 1 Cl, 2 O Balance the K and Cl on the Product Side by adding a coefficient of 2 to KCL. \[2KCLO3-->2KCL+3O2 \] Step 2 Use your periodic table to find out each of the elements atomic weight present in KCLO3 K:39 Cl:35.5 O:16 Molar Mass(g/mol)=(39.00*1)+(35.50*1)+(16.00*3) Use the given mass to plug it in the equation \[Number of Moles=Mass(g)/Molar Mass(g/mol)\] \[Number of Moles of KCLO3=65.2/122.5=0.5322\] Apply the same thing to KCL \[Number of Moles of KCL=39.7/74.5=0.532 \] They Both share the same nunber of moles because their mole ratios are the same but is different when comparing with oxygen gas. The mole ratio of either KCL or KCLO3 to Oxygen is 2:3 \[2:3\] \[0.532:x\] x represent the number of moles of oxygen and can be obtained by cross multiplication method. \[x=0.532*3/2=0.799\] \[Mass=Number of moles*Molar Mass(g/mol)\] \[Mass=0.799*16=12.789 grams\]

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