Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Localized molecular orbitals are molecular orbitals which are concentrated in a limited spatial region of a molecule, for example a specific bond or a lone lake on a specific atom.
1:3 hope DAT helps #ZedTheZom
Answer:
No. Of Moles of zinc = m/Ar
= 13/ 65.38 = 0.198 moles
From balanced equation, Mole ration between CuSO4 and Zn is 1 : 1
So only 0.198 moles of CuSO4 reacts, it is in excess
Mass = no of Moles X Mr
Mass = 0.198 X 159.5 = 31.59 grams
Volume = mass m denisty
Volume j 31.59 / 3.6 = 8.78 ml
Explanation:
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