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alexira [117]
2 years ago
9

How much heat is absorbed/released when 25.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) acc

ording to the following chemical equation? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = 1168 kJ
Chemistry
1 answer:
Novay_Z [31]2 years ago
6 0

Answer:

429.4 kJ are absorbed in the endothermic reaction.

Explanation:

The balanced chemical equation tells us that 1168 kJ of heat are absorbed in the reaction when 4 mol of NH₃ (g) react with 5 mol O₂ (g).

So what we need is to calculates how many moles represent 25 g NH₃(g) and calculate the heat absorbed. (NH₃ is the limiting reagent)

Molar Mass NH₃  = 17.03 g/mol

mol NH₃ = 25.00 g/ 17.03 g/mol = 1.47 mol

1168 kJ /4 mol NH₃  x 1.47 mol  NH₃ =  429.4 kJ

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During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with
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Answer:

8.0 moles

Explanation:

Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.

Using the formula: \frac{concentration of acid X volume of acid}{concentration of base X volume of base} = \frac{mole of acid}{mole of base}

Concentration of acid = ?

Volume of acid = 10 mL

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Volume of base = 40 mL

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Substitute into the equation:

\frac{concentration of acid X 10}{1.0 X 40} = \frac{1}{1}

Concentration of acid = 40/10 = 4.0 M

To determine the number of moles of acid present in 2.0 liters of the unknown solution:

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molarity = 4.0 M

Volume = 2.0 Liters

Hence,

Number of moles = 4.0 x 2.0 = 8 moles

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The three aspects of the measuring process are units, systems, and instruments.
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How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?
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Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
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Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

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