a) $n_S = \dfrac{9,6}{32} = 0,3(mol)$
b) $n_{CO_2} = \dfrac{13,2}{44} = 0,3(mol)$
c) $n_C = \dfrac{1,2}{12} = 0,1(mol)$
d) $n_{C_6H_{12}O_6} = \dfrac{3,6}{180} = 0,02(mol)$
a: \(n=\dfrac{9.6}{32}=0.3\left(mol\right)\)
b: \(n=\dfrac{13.2}{12+16\cdot2}=\dfrac{13.2}{44}=0.3\left(mol\right)\)