Question

Tính KL các ng/tố có trong mỗi lượng chất sau: a,49g axit photphoric. b, 20g sắt(III) oxit. c, 4,48l khí Cacbon đioxit(đktc).

Answer 1

a) \(n_{H_3PO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)

Ta có: \(n_H=3n_{H_3PO_4}=3\times0,5=1,5\left(mol\right)\)

\(\Rightarrow m_H=1,5\times1=1,5\left(g\right)\)

ta có: \(n_P=n_{H_3PO_4}=0,5\left(mol\right)\)

\(\Rightarrow m_P=0,5\times31=15,5\left(g\right)\)

\(\Rightarrow m_O=49-1,5-15,5=32\left(g\right)\)

Answer 2

b) \(m_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\)

Ta có: \(n_{Fe}=2n_{Fe_2O_3}=2\times0,125=0,25\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,25\times56=14\left(g\right)\)

\(\Rightarrow m_O=20-14=6\left(g\right)\)

Answer 3

c) \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Ta có: \(n_C=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_C=0,2\times12=2,4\left(g\right)\)

Ta có: \(n_O=2n_{CO_2}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_O=0,4\times16=6,4\left(g\right)\)