To jest chemia 1. Podręcznik. Zakres podstawowy | Strona 290 | Zadanie 59
Oblicz masę:
a) 2 moli CaO
\[ M_{CaO}\ =\ M_{Ca}\ +\ M_O\ =\ 40\ \frac{g}{mol}\ +\ 16\ \frac{g}{mol}\ =\ \ 56\ \frac{g}{mol}\ \ \]
\[ 1\ mol\ CaO\ -\ 56\ g\]
\[2\ mol\ CaO\ -\ x\ g\]
\[ x\ =\ \frac{2\ mol\ \cdot\ 56\ g}{1\ mol}\ =\ 112\ g\]
b) 1,5 mola H2SO4
\[M_{H_2{SO}_4}=2\ \cdot M_H\ +\ M_S+4\ \cdot M_O\ =2\ \cdot\ 1\ \frac{g}{mol}\ +32\ \frac{g}{mol}\ +\ 4\ \cdot\ 16\ \frac{g}{mol}=\ \ 98\ \frac{g}{mol}\ \ \]
\[ 1\ mol\ H_2{SO}_4\ -\ 98\ g\]
\[ 1,5\ mol\ H_2{SO}_4\ -\ x\ g\]
\[ x\ =\ \frac{1,5\ mol\ \cdot\ 98\ g}{1\ mol}\ =\ 147\ g\]
c) 0,5 mola glukozy C6H12O6
\[ M_{{C_6H}_{12}O_6}=6\ \cdot M_C\ +\ 12\cdot M_H+6\ \cdot M_O\ =6\ \cdot\ 12\ \frac{g}{mol}\ +12\cdot1\ \frac{g}{mol}\ +6\ \cdot\ 16\ \frac{g}{mol}=\ 180\ \frac{g}{mol}\ \ \]
\[ 1\ mol\ {C_6H}_{12}O_6\ -\ 180\ g\]
\[ 0,5\ mol\ {C_6H}_{12}O_6\ -\ x\ g\]
\[x\ =\ \frac{0,5\ mol\ \cdot\ 180\ g}{1\ mol}\ =\ 90\ g\]