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First you take your compound AlCl3 and find the molar mass

26.98gAl+(3)35.45gCl= 133.33gAlCl3

Then you convert moles to grams

5molAlCl3x1g(over)1molx133.33g(over)1g=666.65gAlCl3

if you follow the rule of Significant Figures, 5mol has one significant figure (digit) so your answer needs to have 1 digit

700gAlCl3

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