Pad 3:37 PM * 12%)D, + sjc.cengagenow.com Cengage OWLv2 I Online teaching and le

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Pad 3:37 PM * 12%)D, + sjc.cengagenow.com Cengage OWLv2 I Online teaching and learn... Calculate The Number Of Moles Of G... extra credit" Ch 5 EOC Question 11 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt alculate the number of grams of bromine (Br2) needed to react exactly with 48.9 g of aluminum (Al). The equation for the reaction is Question 2 2A1(s)3Br2(e) -2AlBr3 (s) Question 3 gram Question 4 Question 5 Submit Answer Try Another Version Question 6 5 item attempts remaining Question 7 Visited Question 8 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt 1 pt Question 17 1 pt 1 pt Question 19 1 pt Question 9 Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 18 Progress 6/20 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 Al = 48.9 g / 26.9815 g/mol = 1.81 mole

from the balanced equation we can say that

2 mole of Al requires 3 mole of Br2 so

1.81 mole of Al will require

= 1.81 mole of Al *(3 mole of Br2 / 2 mole of Al)

= 2.72 mole of Br2

mass of 1 mole of Br2 = 159.808 g

so the mass of 2.72 mole of Br2 = 435 g

Therefore, the mass of Br2 required would be 435 g

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