Question

Determine a massa, em gramas, de Al(OH)₃ necessárias para neutralizar 0,05 mol de HCl.?

Pliplis!

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm c\acute alculo~da~massa~molar:}\\\sf m_{Al(OH)_3}=27+3\cdot(16+1)\\\sf m_{Al(OH)_3}=27+51=78~g/mol\\\underline{\rm montagem~da~regra~de~tr\hat es:}\\\sf 1~mol\longrightarrow 78~g\\\sf 0,05\longrightarrow x\\\sf x=78\cdot0,05\\\sf x=3,9~g\end{array}}$