First you initiate by way of calculating the theoretical quantity of CaO produced in this reaction. a million. First convert the two,seven hundred grams of CaCO3 into moles by way of dividing grams by way of molecular weight. you acquire 2700/one hundred.078 (mw of CaCO3) = 26.979 moles. 2. From the equation you will see that that for each mole of CaCO3 you acquire a million mole of CaO as a product. 3. with a view to acquire theoretical grams of CaO produced, rearrange your mole equation to state that grams = moles*molecular weight. 4. Multiply 26.979 moles * fifty six.078 (mw of CaO) to acquire 1512.ninety 3 grams of CaO. 4. p.c. is (unquestionably yield / theoretical yield)*one hundred. making use of this equation you acquire (1060 /1512.ninety 3) * one hundred = 70.06 %.
Answers & Comments
Verified answer
CO2
First you initiate by way of calculating the theoretical quantity of CaO produced in this reaction. a million. First convert the two,seven hundred grams of CaCO3 into moles by way of dividing grams by way of molecular weight. you acquire 2700/one hundred.078 (mw of CaCO3) = 26.979 moles. 2. From the equation you will see that that for each mole of CaCO3 you acquire a million mole of CaO as a product. 3. with a view to acquire theoretical grams of CaO produced, rearrange your mole equation to state that grams = moles*molecular weight. 4. Multiply 26.979 moles * fifty six.078 (mw of CaO) to acquire 1512.ninety 3 grams of CaO. 4. p.c. is (unquestionably yield / theoretical yield)*one hundred. making use of this equation you acquire (1060 /1512.ninety 3) * one hundred = 70.06 %.