The formula of nitrogen oxide is NO, of nitrogen dioxide is NO_2. Write a balanced equation for the reaction of nitrogen oxide with oxygen. For each (1.0) mole of NO: i. how many moles of O_2 are consumed and of NO_2 produced? ii. how many liters of each (state conditions)? iii. how many grams of each? d. Calculate the mass of 4.55 L of HCI gas at STP. e. Find the molecular mass of a gas having a density of a gas having a density of 0.00481 g/mL at STP. f. Write a balanced equation for the reaction you caused to occur in this assignment. Solve the following problems, assume that you used a sample with a mass of 2.550 grams. i. If the sample was pure KCIO_3 and it was completely decomposed into KCI and O_2, fill in the following blanks: There will be ___ moles of O_2 obtained, having a mass of _ grams and occupying _ mL at STP or ___ mL at 29 degree C and 732 torr. ii. Assume now that the sample was an unknown mixture of KCIO_3 with KCI and that complete decomposition yielded 182 mL of oxygen gas measured at 29 degree C and 732 torr. Calculate the percent by mass of KCIO_3 in the mixture. g. Nitrogen gas was collected over water at 26 degree C and 755 torr. If the volume collected was 35.9L. What volume would it occupy, dry, at STP?

Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) The mass of each compound are:

1 mol NO×[tex]\frac{30 g}{1molNO}[/tex] = 30g

0,5 mol O₂×[tex]\frac{32 g}{1molO_{2}}[/tex] = 16g

1 mol NO₂×[tex]\frac{46 g}{1molNO_{2}}[/tex] = 46g

d) Using:

n = PV / RT

Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:

0,203 moles of HCl. In grams:

0,203 mol HCl×[tex]\frac{36,46 g}{1molHCl}[/tex] = 7,41 g of HCl

e) Using:

δRT/P = MW

Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)

And MW is molecular mass: 107,7 g/mol

f) For the reaction:

2 KClO₃ → 2 KCl + 3 O₂

2,550 g of KClO₃ are:

2,550 g of KClO₃×[tex]\frac{1mol}{122,55 gKClO_{3}}[/tex] = 0,0208 moles of KClO₃

When these moles reacts completely produce:

0,0208 moles of KClO₃×[tex]\frac{3 mol O_{2}}{2 molKClO_{3}}[/tex] = 0,0312 moles of O₂

In grams:

0,0312 moles of O₂ ×[tex]\frac{32g}{1 molO_{2}}[/tex] = 0,999g of O₂

V = nRT/P

At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; V = 0,699L ≡ 699mL

At 29 °C (302,15K) and 732 torr (0,963 atm)

V = 0,803L ≡ 803mL

ii. 182 mL ≡ 0,182L of O₂ are:

n = PV/RT

moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:

7,07x10⁻³ moles of O₂×[tex]\frac{2 mol KClO_{3}}{3 molO_{2}}[/tex] = 0,0106 mol KClO₃. In grams:

0,0106 moles of KClO₃×[tex]\frac{122,55 g}{1molKClO_{3}}[/tex] = 1,300 g of KClO₃.

Thus, percent by mass of KClO₃ in the mixture is:

1,300g/2,550g ×100 = 51%

g. Combined gas law says that:

[tex]\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}[/tex]

Where:

P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) V₂ = 32,6 L

I hope it helps!