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) __ =

12. To calculate the moles of NO formed, first determine the moles of \( NH_3 \) used:<br /><br />- Molar mass of \( NH_3 = 14.01 + (3 \times 1.01) = 17.04 \, \text{g/mol} \)<br />- Moles of \( NH_3 = \frac{824 \, \text{g}}{17.04 \, \text{g/mol}} \approx 48.35 \, \text{mol} \)<br /><br />From the balanced equation, 4 moles of \( NH_3 \) produce 4 moles of \( NO \). Therefore, the moles of \( NO \) produced are equal to the moles of \( NH_3 \):<br /><br />- Moles of \( NO = 48.35 \, \text{mol} \)<br /><br />13. Using the same reaction and moles of \( NH_3 \), calculate the moles of water (\( H_2O \)) produced:<br /><br />From the balanced equation, 4 moles of \( NH_3 \) produce 6 moles of \( H_2O \). Therefore, the moles of \( H_2O \) produced are:<br /><br />- Moles of \( H_2O = \left(\frac{6}{4}\right) \times 48.35 \, \text{mol} = 72.525 \, \text{mol} \)<br /><br />14. (A) Calculate moles of \( HgO \) needed to produce 125 g of \( O_2 \):<br /><br />- Molar mass of \( O_2 = 32.00 \, \text{g/mol} \)<br />- Moles of \( O_2 = \frac{125 \, \text{g}}{32.00 \, \text{g/mol}} \approx 3.906 \, \text{mol} \)<br /><br />From the balanced equation, 1 mole of \( O_2 \) is produced from 2 moles of \( HgO \). Therefore, the moles of \( HgO \) needed are:<br /><br />- Moles of \( HgO = 2 \times 3.906 \, \text{mol} = 7.812 \, \text{mol} \)<br /><br />(B) Calculate moles of mercury (\( Hg \)) produced when 125 g of \( O_2 \) is also produced:<br /><br />From the balanced equation, 1 mole of \( O_2 \) produces 2 moles of \( Hg \). Therefore, the moles of \( Hg \) produced are:<br /><br />- Moles of \( Hg = 2 \times 3.906 \, \text{mol} = 7.812 \, \text{mol} \)