Pad? 4:03 PM * 14% + sjc.cengagenow.com Wendy Avila OWLv2 I Online teaching and

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Pad? 4:03 PM * 14% + sjc.cengagenow.com Wendy Avila OWLv2 I Online teaching and learn... A Product Weighing 14.44 G Was Isol... extra credit" Ch 5 EOC Question 11 pt 1 pt 1 pt 1 pt Question 5 1 pt 1 pt Question 71 pt 1 pt 1 pt A sample of calcium metal with a mass of 2.13 g was reacted with excess oxygen. The following equation represents the reaction that took place Question 2 2Ca(s) + O2(g) -2CaO (s) Question 3 The isolated product (CaO) weighed 2.18 g. What was the percentage vield of the reaction? Question 4 Percentage yield- Question 6 Submit Answer Try Another Version 5 item attempts remaining Question 8 Question 9 Question 10 Question 11 1 pt Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 1 pt Question 18 Question 19 1 pt 1 pt 1 pt 1 pt Visited Progress 10/20 items Due Apr 1 at 11:55 PM Previous Next Finish Assignment Save and Exit Cengage Learning| Cengage Technical Support

Explanation / Answer

Number of moles of Ca = 2.13 g / 40.078 g/mol = 0.0531 mole

From the balanced equation we can say that

2 mole of Ca produces 2 mole of CaO so

0.0531 mole of Ca will produce

= 0.0531 mole of Ca *(2 mole of CaO / 2 mole of Ca)

= 0.0531 mole of CaO

mass of 1 mole of CaO = 56.077 g

so the mass of 0.0531 mole of CaO = 2.98 g

Therefore, theoretical yield of CaO = 2.98 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (2.18 / 2.98)*100 = 73.2 %

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