Question

nung a gam KMnO4 thu được lượng oxi vừa đủ oxi hóa 7,84 gam S    

a) tìm a

 

Answer 1

\(n_S=\dfrac{7.84}{32}=0.245\left(mol\right)\)

\(2KMnO_4\underrightarrow{^{t^0}}K_2MnO_4+MnO_2+O_2\)

\(S+O_2\underrightarrow{^{t^0}}SO_2\)

\(n_{KMnO_4}=2n_{O_2}=2n_S=2\cdot0.245=0.49\left(mol\right)\)

\(m_{KMnO_4}=0.49\cdot158=77.42\left(g\right)\)

Answer 2

\(S + O_2 \xrightarrow{t^o} SO_2\\ n_{O_2} = n_S = \dfrac{7,84}{32}= 0,245(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,49(mol)\\ a = 0,49.158 = 77,42(gam)\)