1.133 g of silver nitrate was heated in an open tube. The silver residue weighed 0.720 g. During the reaction, 0.307 g of nitrogen dioxide was also produced. The rest of the mass loss was due to oxygen. Use the data to write the equation for the reaction.
Answer (3)
Ag_2O+NO_2\uparrow+O_2\uparrow\\\\ N^{+V}+1e\rightarrow N^{+IV}|4\\ 2O^{-II}-4e\rightarrow O_2^0|1\\\\ 4AgNO_3--^T-->2Ag_2O+4NO_2\uparrow + O_2\uparrow"> A g N O 3 − − T − − > A g 2 O + N O 2 ↑ + O 2 ↑ N + V + 1 e → N + I V ∣4 2 O − II − 4 e → O 2 0 ∣1 4 A g N O 3 − − T − − > 2 A g 2 O + 4 N O 2 ↑ + O 2 ↑
The equation for the **reaction **is . 2 A g N O 3 →2Ag + 2 N O 2 + O 2 ;
The balanced equation for the decomposition of silver nitrate is 2AgNO₃(s) → 2Ag(s) + 2NO₂(g) + O₂(g). This equation indicates that silver nitrate decomposes upon heating, resulting in the formation of silver, nitrogen dioxide, and oxygen. Each part of the reaction reflects the conservation of mass based on the data provided.
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