Balance the chemical equation given below,and calculate the volume of nitrogen monoxide gas produced when 8.00 g of ammonia is reacted with 12.0 g of oxygen at 25 °C.The density of nitrogen monoxide at $25 ^ { \circ } \mathrm { C }$ is 1.23 g L^-1. ________ NH₃(g)+ ________ O₂(g)? ________ NO(g)+ ________ H₂O(l)

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The balanced chemical equation is:4NH3(g)+ 5O2(g)→ 4NO(g)+ 6H2O(l)To calculate the volume of nitrogen monoxide gas produced, we need to first calculate the moles of ammonia and oxygen used in the reaction.Moles of ammonia = 8.00 g / 17.03 g/mol = 0.470 molMoles of oxygen = 12.0 g / 32.00 g/mol = 0.375 molFrom the balanced chemical equation, we can see that 4 moles of ammonia react with 5 moles of oxygen. Therefore, the limiting reactant in this reaction is oxygen.Moles of nitrogen monoxide produced = 0.375 mol * (4 mol NO / 5 mol O2) = 0.300 molVolume of nitrogen monoxide produced = 0.300 mol * 22.4 L/mol = 6.72 LHowever, the density of nitrogen monoxide at 25 °C is 1.23 g L-1. Therefore, the volume of nitrogen monoxide gas produced at 25 °C is:Volume of nitrogen monoxide produced = 6.72 L * (1.23 g L-1 / 1 g L-1) = 8.26 LTherefore, the correct answer is A) 7.32 L.

Balance the Chemical Equation Given Below,and Calculate the Volume of Nitrogen