Hóa học tính khối lượng của các nguyên tố có trong:4(g) NaOH; 32g Fe2(SO4)3 04/09/2021 By Skylar tính khối lượng của các nguyên tố có trong:4(g) NaOH; 32g Fe2(SO4)3
Xét `4g NaOH` `n_(NaOH)=\frac{4}{40}=0,1(mol)` `=>n_(Na)=0,1(mol)` `=>m_(Na)=0,1.23=2,3(g)` `n_(O)=0,1(mol)` `=>m_(O)=0,1.16=1,6(g)` `m_(H)=0,1.1=0,1(g)` Xét `32g` `Fe_2(SO_4)_3` `n_(Fe_2(SO_4)_3)=\frac{32}{400}=0,08(mol)` `n_(Fe)=0,16(mol)` `m_(Fe)=0,16.56=8,96(g)` `n_(S)=0,08.3=0,24(mol)` `m_(S)=0,24.32=7,68(g)` `n_(O)=0,08.12=0,96(g)` `m_(O)=0,96.16=15,36(g)` Trả lời
Xét 4 gam \(NaOH\). \({n_{NaOH}} = \frac{4}{{23 + 16 + 1}} = 0,1{\text{ mol = }}{{\text{n}}_{Na}} = {n_O} = {n_H}\) \( \to {m_{Na}} = 0,1.23 = 2,3{\text{ gam}}\) \({m_O} = 0,1.16 = 1,6{\text{ gam}}\) \({m_H} = 0,1.1 = 0,1{\text{ gam}}\) Xét 32 gam \(Fe_2(SO_4)_3\). Ta có: \({n_{F{e_2}{{(S{O_4})}_3}}} = \frac{{32}}{{56.2 + (32 + 16.4).3}} = 0,08{\text{ mol}}\) \( \to {n_{Fe}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,08.2 = 0,16{\text{ mol}}\) \({n_S} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,24{\text{ mol}}\) \({n_O} = 12{n_{F{e_2}{{(S{O_4})}_3}}} = 0,96{\text{ mol}}\) \( \to {m_{Fe}} = 0,16.56 = 8,96{\text{ gam}}\) \( \to {m_S} = 0,24.32 = 7,68{\text{ gam}}\) \( \to {m_O} = 0,96.16 = 15,36{\text{ gam}}\) Trả lời
Xét `4g NaOH`
`n_(NaOH)=\frac{4}{40}=0,1(mol)`
`=>n_(Na)=0,1(mol)`
`=>m_(Na)=0,1.23=2,3(g)`
`n_(O)=0,1(mol)`
`=>m_(O)=0,1.16=1,6(g)`
`m_(H)=0,1.1=0,1(g)`
Xét `32g` `Fe_2(SO_4)_3`
`n_(Fe_2(SO_4)_3)=\frac{32}{400}=0,08(mol)`
`n_(Fe)=0,16(mol)`
`m_(Fe)=0,16.56=8,96(g)`
`n_(S)=0,08.3=0,24(mol)`
`m_(S)=0,24.32=7,68(g)`
`n_(O)=0,08.12=0,96(g)`
`m_(O)=0,96.16=15,36(g)`
Xét 4 gam \(NaOH\).
\({n_{NaOH}} = \frac{4}{{23 + 16 + 1}} = 0,1{\text{ mol = }}{{\text{n}}_{Na}} = {n_O} = {n_H}\)
\( \to {m_{Na}} = 0,1.23 = 2,3{\text{ gam}}\)
\({m_O} = 0,1.16 = 1,6{\text{ gam}}\)
\({m_H} = 0,1.1 = 0,1{\text{ gam}}\)
Xét 32 gam \(Fe_2(SO_4)_3\).
Ta có:
\({n_{F{e_2}{{(S{O_4})}_3}}} = \frac{{32}}{{56.2 + (32 + 16.4).3}} = 0,08{\text{ mol}}\)
\( \to {n_{Fe}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,08.2 = 0,16{\text{ mol}}\)
\({n_S} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,24{\text{ mol}}\)
\({n_O} = 12{n_{F{e_2}{{(S{O_4})}_3}}} = 0,96{\text{ mol}}\)
\( \to {m_{Fe}} = 0,16.56 = 8,96{\text{ gam}}\)
\( \to {m_S} = 0,24.32 = 7,68{\text{ gam}}\)
\( \to {m_O} = 0,96.16 = 15,36{\text{ gam}}\)