Based on the balanced equation 2NBr 3 + 3NaOH N 2 + 3NaBr + 3HOBr calculate the

ID: 580300 • Letter: B

Question

Based on the balanced equation

2NBr3 + 3NaOH N2 + 3NaBr + 3HOBr

calculate the number of HOBr molecules formed when 44 NBr3 molecules and 96 NaOH formula units react?

6.022×1023 mol-1

Based on the balanced equation

Molar Mass (g/mol) NBr3 253.72 NaOH 39.997 N2 28.013 NaBr 102.800 HOBr 96.911 Avogadro's No.

6.022×1023 mol-1

Based on the balanced equation

Mg + Cl2 MgCl2

calculate the number of MgCl2 formula units formed when 73 Mg atoms and 72 Cl2 molecules react?

Molar Mass (g/mol) Mg 24.305 Cl2 70.906 MgCl2 95.211 Avogadro's No. 6.022×1023 mol-1

2NBr3 + 3NaOH N2 + 3NaBr + 3HOBr

First calculate the limiting reagent.

2 molecules of NBr3 reacts with 3 moles of NaOH

44 molecules of NBr3 reacts with 44 x 3 / 2 = 66 molecules of NaOH

but we have 96 molecules of NaOH so NaOH is exess reagent.

NBr3 is limiting reagent.

now

2 molecules of NBr3 forms 3 molecules of HOBr

44 molecules of NBr3 forms 44 x 3 / 2 = 66

molecules of HOBr formed = 66

2) Mg + Cl2 -----------> MgCl2

according to balanced reaction

Cl2 is limiting reagent.

1 molecule Cl2 forms 1 molecule MgCl2

72 molecules forms 72 x 1 / 1 = 72

molecules of MgCl2 forms = 72

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