Question

Answer 1

1:

a) 4Al +3 O₂ --t^o--> 2Al₂O₃

b) Theo ĐLBTKL: m_Al + m_O2 = m_Al2O3 (1)

c) (1) => m_O2 = 20,4-10,8=9,6(g)

2:

a) Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

b) Theo ĐLBTKL: m_Na2CO3 + m_HCl = m_NaCl +m_CO2 + m_H2O (1)

c) (1) => m_HCl = 11,7 + 1,8 + 4,4 - 10,6 = 7,3 (g)

Answer 2

\(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)

\(\text{Áp dụng ĐLBTKL, ta có:}\text{m}_{Al}\)\(+m_{O_2}\)\(=m_{Al_2}\)\(_{O_3}\)

\(\Rightarrow m_{O_2}\)\(=20,4-10,8=9,6\left(g\right)\)