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?

a) 0,5 mol O₂; 1 mol NO₂

b)

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) NO = 30g

O₂ = 16g

NO₂ = 46g

d) 7,41g HCl

e) 107,7 g/mol

f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL

ii. 51%

g) 32,6L

Explanation:

a) For the reaction:

2 NO + O₂ → 2 NO₂

For 1 mole of NO there are consumed:

1 mol NO × = 0,5 mol of O₂

And produced:

1 mol NO × = 1 mol of NO₂

b) By ideal gas law:

V = nRT/P

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× = 30g

0,5 mol O₂× = 16g

1 mol NO₂× = 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× = 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× = 0,0208 moles of KClO₃

When these moles reacts completely produce:

0,0208 moles of KClO₃× = 0,0312 moles of O₂

In grams:

0,0312 moles of O₂ × = 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× = 0,0106 mol KClO₃. In grams:

0,0106 moles of KClO₃× = 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:

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

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