Question
$$27 g$$ of $$Al$$ will react completely with how many grams of oxygen?
A.
$$8\,g$$
B.
$$16\,g$$
C.
$$32\,g$$
D.
$$24\,g$$
Answer :
$$24\,g$$
Solution :
$$\eqalign{
& 4Al + 3{O_2} \to 2A{l_2}{O_3} \cr
& {\text{At}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of}}\,Al = 27 \cr
& {\text{Thus}}\,4 \times 27g\,{\text{of}}\,Al\,{\text{reacts}}\,{\text{with}}\,{\text{oxygen}} = 3 \times 32g \cr
& \therefore 27{\text{g}}\,{\text{of}}\,Al\,{\text{reacts}}\,{\text{with}}\,{\text{oxygen}} = \frac{{3 \times 32}}{{4 \times 27}} \times 27g = 24g \cr} $$