Question

balance the chemical equation given below and determine the moles of iodine that reacts with 10g of aluminium.
Al(s) + I(s) = Al2I6

Answer 1

ANSWER = 1.11 mole

$Al_{(s)} + I_{(s)} \rightarrow Al_{2}I_{6}$

Balanced Form ::

$2Al_{(s)} +6 I_{(s)} \rightarrow Al_{2}I_{6}$

Now calculating the amount of I required

$2 \: mol \: Al \implies 6 \: mol \: I$

$2 × 27 g \: Al \implies 6 × 127 \: g \: I$

$for \: 10 g \: Al \implies \frac{ 6 × 127 × 10 }{2× 27} g \: \: I$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 141.11g \: I$

$no. \: of \: moles \: of \: I = \frac{141.11}{127}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1.11 \: mole$