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Sagot :
First, we are going to write the reaction for this.
We need to balance the reaction too.
[tex]\begin{gathered} 2Al(s)+6HBr(g)======>2AlBr_3(s)+3H_2(g) \\ \end{gathered}[/tex]-----------------------------------------------------------------
2Al + 6 HBr ===> 2AlBr3 + 3H2
15.g x
We call x the grams of HBr that are going to react with 15.0 g of HBr
----------------------------------------------------------------
The atomic mass of Al = 26.98 g/mol (from the periodic table)
(1 mol of Al = 26.98 g)
The molecular mass of HBr = 80.91 g/mol
(1 mol of HBr = 80.91 g)
---------------------------------------------------------------
2Al + 6 HBr ===> 2AlBr3 + 3H2
15.g x
2(moles) x 26.98 g of Al ----------------------------6 (moles) x 80.91 g of HBr
15.0 g of Al ----------------------------- x
[tex]x\text{ = }\frac{15.0\text{ g x 6 x 80.91 g}}{2x26.98\text{ g}}=\text{ 134.94 g of HBr}[/tex]------------------------------------------------------------
Answer: grams of HBr=134.94 g
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