Question

If analytical chemist found approximately 12 moles of oxygen in a sample of (Al${}_2$O${}_3$) and she also knows that the atomic mass of aluminum is 27 g/mol and the atomic mass of oxygen is 16 g/mol then what total mass of this Al${}_2$O${}_3$ is :

• A. 102 g
• B. 192 g
• C. 3672 g
• D. 408 g
• E. 1224 g

Answer 1

Answer: (D)

408 g

Total molecular mass of $A{l}_2{O}_3$ $=27\times 2+16\times 3=102gm$

$3$ moles of oxygen are given by one mole of $A{l}_2{O}_3$

hence $12$ moles of oxygen is given by $4$ moles of $A{l}_2{O}_3$

$1$ mole of $A{l}_2{O}_3$ gives $16\times 3=102gm$

$4$ moles of $A{l}_2{O}_3=102\times 4=408g.$