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12. The First Step in the Industrial Manufacturing of Nitric Acid Is the Catalytic Oxidation of Ammonia. 4NH_(3(g))+5O_(2(g))arrow

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12. The first step in the industrial manufacturing of nitric acid is the catalytic oxidation of ammonia. 4NH_(3(g))+5O_(2(g))arrow 4NO_((g))+6H_(2)O_((g)) The reaction is run using 824g of NH_(3) and excess oxygen, calculate how many moles of NO are formed. 13. If the same reaction from# 12 is run using 824 gof NH_(3) and excess oxygen, calculate the molar quantity of water produced. (Excess exypen you won't run out of O_(2) The reoction will be complete when you use up oll the NH_(2) __ 14. Oxygen was discovered by Joseph Priestly in 1774 when he heated mercury(II)oxide and it decomposed into mercury and oxygen. 2HgO_((s))arrow 2Hg_((s))+O_(2(g)) (A) Calculate how many moles of mercury(II) oxide, HgO, are needed to produce 125 g of oxygen, O_(2) (B) Calculate how many moles of mercury are p produced when you also produce 125 gof O_(2) __ =

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Ángel professionell · Tutor durante 6 años
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12. To calculate the moles of NO formed, first determine the moles of used:- Molar mass of \( NH_3 = 14.01 + (3 \times 1.01) = 17.04 \, \text{g/mol} \)- Moles of From the balanced equation, 4 moles of produce 4 moles of . Therefore, the moles of produced are equal to the moles of :- Moles of 13. Using the same reaction and moles of , calculate the moles of water ( ) produced:From the balanced equation, 4 moles of produce 6 moles of . Therefore, the moles of produced are:- Moles of \( H_2O = \left(\frac{6}{4}\right) \times 48.35 \, \text{mol} = 72.525 \, \text{mol} \)14. (A) Calculate moles of needed to produce 125 g of :- Molar mass of - Moles of From the balanced equation, 1 mole of is produced from 2 moles of . Therefore, the moles of needed are:- Moles of (B) Calculate moles of mercury ( ) produced when 125 g of is also produced:From the balanced equation, 1 mole of produces 2 moles of . Therefore, the moles of produced are:- Moles of