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汽车剧烈碰撞时,安全气囊中发生反应: $10 \mathrm{NaN}_{3}+2 \mathrm{KNO}_{3}=\mathrm{K}_{2} \mathr...
Mar 19, 2024
汽车剧烈碰撞时,安全气囊中发生反应: 10NaN3+2KNO3=K2O+5Na2O+16 N210 \mathrm{NaN}_{3}+2 \mathrm{KNO}_{3}=\mathrm{K}_{2} \mathrm{O}+5 \mathrm{Na}_{2} \mathrm{O}+16 \mathrm{~N}_{2} \uparrow 。若氧化产物比还原产物多 1.75 mol1.75 \mathrm{~mol} ,则下列判断不正确的是 A. 生成44.8 LN2\mathrm{L} \mathrm{N}_{2} (标准状况) B. 有 0.25 molKNO0.25 \mathrm{~mol} \mathrm{KNO} 被还原 C. 转移电子的物质的量为 1.75 mol1.75 \mathrm{~mol} D. 被氧化的 NN 原子的物质的量为 3.75 mol3.75 \mathrm{~mol}
A
Key Concept
Stoichiometry and gas laws
Explanation
The stoichiometry of the reaction does not support the formation of 44.8 L of N2 N_2 gas under standard conditions (22.4 L/mol). The correct volume should be calculated based on the stoichiometry of the balanced equation.
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