Question

2) Qual o volume ocupado nas CNTP, por 85 g de gás amônia (NH)? Dados: N = 14
e H = 1
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Answer 1

$\large\boxed{\begin{array}{l}\sf A~formula~da~am\hat onia~\acute e~NH_3.\\\sf m_{NH_3}=14+3\cdot1=17~g/mol\\\sf 17~g\longrightarrow 22,4L\\\sf 85~g\longrightarrow x\\\sf 17\cdot x=85\cdot22,4\\\sf x=\dfrac{\diagdown\!\!\!\!\!\!85^5\cdot22,4}{\diagdown\!\!\!\!\!\!17_1}\\\sf x=5\cdot22,4\\\sf x=112~L\end{array}}$

Answer 2

Massa molar NH3 = 14 + 1 x 3 = 17 g/mol

número de mols:

85 / 17 => 5,0 mols

finalizando..

1 mol -------------- 22,4 L ( nas CNTP)

5,0 mols ------------x L

x = 5,0 . 22,4

x = 112 L