Question

dans la réaction suivante :
2 al + 3 cl₂ → 2 alcl₃
calculez la masse de chlorure d’aluminium (alcl₃) produite à partir de 54 g d’aluminium (al).
5 points
a 133 g
b 152 g
c 266 g
d 897 g
e 333 g
f 450 g

Explanation:

Step1: Calculate moles of Al

Molar mass of Al is \(27\space g/mol\). Moles of Al = \(\frac{mass}{molar\space mass}=\frac{54\space g}{27\space g/mol} = 2\space mol\).

Step2: Relate moles of Al and \(AlCl_3\)

From the reaction \(2\space Al + 3\space Cl_2
ightarrow2\space AlCl_3\), the mole ratio of \(Al\) to \(AlCl_3\) is \(2:2 = 1:1\). So moles of \(AlCl_3\) = moles of Al = \(2\space mol\).

Step3: Calculate molar mass of \(AlCl_3\)

Molar mass of \(AlCl_3\) = \(27 + 3\times35.5 = 27 + 106.5 = 133.5\space g/mol\).

Step4: Calculate mass of \(AlCl_3\)

Mass = moles \(\times\) molar mass = \(2\space mol\times133.5\space g/mol = 267\space g\) (approx \(266\space g\) due to rounding in molar mass of Cl as \(35.5\)).

Answer:

C. 266 g