a) \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
Ta có: \(n_{Fe}=2n_{Fe_2O_3}=2\times0,2=0,4\left(mol\right)\)
\(n_O=3n_{Fe_2O_3}=3\times0,2=0,6\left(mol\right)\)
Số nguyên tử Fe là: \(0,4\times6\times10^{23}=2,4\times10^{23}\left(ngtử\right)\)
Số nguyên tử O là: \(0,6\times6\times10^{23}=3,6\times10^{23}\left(ngtử\right)\)
b) \(n_{C_{12}H_{22}O_{11}}=\dfrac{34,3}{342}=\dfrac{343}{3420}\left(mol\right)\)
Ta có: \(n_C=12n_{C_{12}H_{22}O_{11}}=12\times\dfrac{343}{3420}=\dfrac{343}{285}\left(mol\right)\)
\(n_H=22n_{C_{12}H_{22}O_{11}}=22\times\dfrac{343}{3420}=\dfrac{3773}{1710}\left(mol\right)\)
\(n_O=11n_{C_{12}H_{22}O_{11}}=11\times\dfrac{343}{3420}=\dfrac{3773}{3420}\left(mol\right)\)
Số nguyên tử C là: \(\dfrac{343}{285}\times6\times10^{23}=7,221052632\times10^{23}\left(ngtử\right)\)
Số nguyên tử H là: \(\dfrac{3773}{1710}\times6\times10^{23}=1,323859649\times10^{24}\left(ngtử\right)\)
Số nguyên tử O là: \(\dfrac{3773}{3420}\times6\times10^{23}=6,619298246\times10^{23}\left(ngtử\right)\)
c) \(n_{Cl_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Ta có: \(n_{Cl}=2n_{Cl_2}=2\times0,6=1,2\left(mol\right)\)
Số nguyên tử Cl là: \(1,2\times6\times10^{23}=7,2\times10^{23}\left(ngtử\right)\)
d) \(n_{CH_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Ta có: \(n_C=n_{CH_4}=0,3\left(mol\right)\)
\(n_H=4n_{CH_4}=4\times0,3=1,2\left(mol\right)\)
Số nguyên tử C là: \(0,3\times6\times10^{23}=1,8\times10^{23}\left(ngtử\right)\)
Số nguyên tử H là: \(1,2\times6\times10^{23}=7,2\times10^{23}\left(ngtử\right)\)
