Question

Tính khối lượng các nguyên tố có trong 11.5 gam C₂H₆O

Answer 1

\(n_{C_2H_6O}=\dfrac{11,5}{46}=0,25\left(mol\right)\\ m_C=0,25.2.12=6\left(g\right)\\ m_H=6.0,25.1=1,5\left(g\right)\\ m_O=0,25.1.16=4\left(g\right)\)

Answer 2

\(n_{C_2H_6O}=\dfrac{m}{M}=\dfrac{11.5}{46}=0,25\left(mol\right)\)

\(\left\{{}\begin{matrix}n_C=0,25.2=0,5\left(mol\right)\\n_H=0,2.6=1,5\left(mol\right)\\n_O=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_C=n.M=0,5.12=6\left(g\right)\\m_H=n.M=1,5.1=1,5\left(g\right)\\m_O=n.M=0,25.16=4\left(g\right)\end{matrix}\right.\)

Answer 3

\(\%m_C=\dfrac{2.12.100\%}{46}=52,17\%\Rightarrow m_C=52,17\%.11,5=6\left(g\right)\)

\(\%m_H=\dfrac{6.1.100\%}{46}=13,04\%\Rightarrow m_H=13,04\%.11,5=1,5\left(g\right)\)

\(m_O=11,5-6-1,5=4\left(g\right)\)