Question

3. Converta as seguintes massas em quantidades (em mols) e número de moléculas (ou átomos) a) 240 g de alumínio em átomos(Al) ( A1=27) b) 10 kg de KNO3 (K-39, N = 14 O=16) c) 0,37 g de N*H_{3}(N = 14; H = 1).​

Answer 1

$\Large\boxed{\begin{array}{l}\rm a )\\\sf 27\,g\longrightarrow 6\cdot10^{23}\,\acute atomos\\\sf 240\,g\longrightarrow x\\\sf 27x=240\cdot6\cdot10^{23}\\\\\sf x=\dfrac{1440\cdot10^{23}}{27}\\\\\sf x=53,33\cdot10^{23}=5,33\cdot10^{24}\,\acute atomos\end{array}}$

$\Large\boxed{\begin{array}{l}\rm b)\\\sf m_{KNO_3}=39+14+3\cdot16=101\,g/mol\\\sf 10\,kg=10\cdot10^3=10^4\,g\\\sf 101 \,g\longrightarrow 6\cdot10^{23}\,\acute atomos\\\sf 10^4\,g\longrightarrow x\\\sf 101x=6\cdot10^{23}\cdot10^4\\\sf x=\dfrac{6\cdot10^{27}}{101}\\\\\sf x=0,05\cdot10^{27}\\\sf x=5\cdot10^{25}\,\acute atomos\end{array}}$

$\Large\boxed{\begin{array}{l}\rm b)\\\sf m_{NH_3}=14+3\cdot1=14+3=17\,g/mol\\\sf 17\,g\longrightarrow 6\cdot10^{23}\,\acute atomos\\\sf 0,37\,g\longrightarrow x\\\sf 17x=6\cdot0,37\cdot10^{23}\\\\\sf x=\dfrac{2,22\cdot10^{23}}{17}\\\\\sf x=0,13\cdot10^{23}\\\sf x=1,3\cdot10^{22}\,\acute atomos\end{array}}$