Question

Bài 2: Tìm số mol nguyên tử của mỗi nguyên tố có trong:
1. 17 gam NH₃
2. 1,7 gam AgNO₃
3. 13,3 gam (NH₄)₂SO₄
4. 2,94 gam K₂Cr₂O₇

Answer 1

1.

\(m_N=\dfrac{14.17}{14+3}=14\left(g\right)\Rightarrow n_N=1\left(mol\right)\)

\(m_H=\dfrac{3.17}{14+3}=3\left(g\right)\Rightarrow n_H=3\left(mol\right)\)

2.

\(m_{Ag}=\dfrac{108.1,7}{108+14+16.3}=1,08\left(g\right)\Rightarrow n_{Ag}=0,01\left(mol\right)\)

\(m_N=\dfrac{14.1,7}{108+14+16.3}=0,14\left(g\right)\Rightarrow n_N=0,01\left(mol\right)\)

\(m_O=\dfrac{16.3.1,7}{108+14+16.3}=0,48\left(g\right)\Rightarrow n_O=0,03\left(mol\right)\)

Answer 2

3.

\(m_N=\dfrac{14.2.13,3}{\left(14+4\right).2+32+16.4}=2,82\left(g\right)\Rightarrow n_N=0,2\left(mol\right)\)

\(m_H=\dfrac{4.2.13,3}{\left(14+4\right).2+32+16.4}=0,81\left(g\right)\Rightarrow n_H=0,81\left(mol\right)\)

\(m_S=\dfrac{32.13,3}{\left(14+4\right).2+32+16.4}=3,22\left(g\right)\Rightarrow n_S=0,1\left(mol\right)\)

\(m_O=\dfrac{16.4.13,3}{\left(14+4\right).2+32+16.4}=6,45\left(g\right)\Rightarrow n_O=0,4\left(mol\right)\)

Answer 3

4.

\(m_K=\dfrac{39.2.2,94}{39.2+52.2+16.7}=0,78\left(g\right)\Rightarrow n_K=0,02\left(mol\right)\)

\(m_{Cr}=\dfrac{52.2.2,94}{39.2+52.2+16.7}=1,04\left(g\right)\Rightarrow n_{Cr}=0,02\left(mol\right)\)

\(m_O=\dfrac{16.7.2,94}{39.2+52.2+16.7}=1,12\left(g\right)\Rightarrow n_O=0,07\left(mol\right)\)