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Misha Larkins [42]
3 years ago
13

The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)

, ΔHrxn=−290.0 kJ Find the change in enthaply when 20.0 g of NI3 decomposes.
Chemistry
1 answer:
EastWind [94]3 years ago
3 0

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = \frac{20.0 g}{395 g/mol}=0.05063 mol

2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

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A 0.00143 M  concentration of MnO4^- is not a reasonable solution .

<h3>Number of moles of carbonate</h3>

The ions left in solution are Na^+ and NO3^-

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Since;

1 mole of sodium carbonate reacts with 1 mole of calcium nitrate  then 0.1 moles of sodium carbonate were used.

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The claim of the student that the concentration of sodium carbonate is too low is wrong because the value was calculated from concentration  and volume of calcium nitrate  and not using the precipitate. If the filtrate is tested for conductivity, it will be found to conduct electricity because it contains sodium and NO3 ions.

2) In the reaction as shown, the MnO4^- ion was reduced.

The initial volume is 3.4 mL while the final volume is 29.6 mL.

Number of moles of MnO4^- ion = (29.6 mL - 3.4 mL)/1000 × 0.0235 M = 0.0006157 moles

<h3>The calculations are performed as follows</h3>

  • If 2 moles of MnO4^- reacted with 5 moles of acid

0.0006157 moles of MnO4^- reacted with  0.0006157 moles ×  5 moles/ 2 moles

= 0.0015 moles

  • In this case, number of moles of acid = 0.139 g/90 g/mol = 0.0015 moles

Number of moles of MnO4^-  = 0.00143 M × (29.6 mL - 3.4 mL)/1000

= 0.000037 moles

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0.000037 moles of MnO4^- reacts with 0.000037 moles × 5 moles/ 2 moles

= 0.000093 moles

  • Hence, this is not a reasonable amount of solution.

Learn more about MnO4^- : brainly.com/question/10887629

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Answer:

See explanations

Explanation:

Stoichiometry is very easy to master if you understand the ‘mole concept’ and how it is used to define and describe chemical process mathematically. A ‘mole’ – in chemistry – is the mass of substance containing one Avogadro’s Number of particles. That is, N₀ = 6.023 x 10²³ particles / mole. When working with chemical reactions and equations data should be first converted to moles using the following conversations:

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In this problem you are given the equation Na + H₂O => NaOH + H₂. ‘Reading the equation’ there is 1 mole of Na, 1 mole of water, 1 mole of NaOH and 1 mole of H₂. In another example 3H₂ + N₂ => 2NH₃ there are 3 moles of H₂, 1 mole of N₂ and 2 moles of NH₃. The mole values can be multiples or fractions but if one mole value increases all the remaining mole values increase or decrease proportionally. For example:

Using the equation Na + H₂O => NaOH + H₂, one could apply a 2 before the Na but all the following formulas would need be increased by a factor of 2. If one applies ½ to the Na then all the following formulas would need be cut in half also and the reaction stoichiometry would still be valid. The fact that the equation is written with coefficients of 1 is that it is in the smallest whole number ratio of coefficients. This then implies the reaction formula is in ‘standard form’. This also implies the equation conditions are at 0⁰C & 1atm pressure and 1 mole of any gas phase substance occupies 22.4 Liters volume. Such is the significance of converting given data to moles as all other substance mass (in moles) are proportional.  

For your 1st problem, 1.76 x 10²⁴ formula units of Na will react with water (usually read as an excess) to produce (?) grams of H₂.

1st write the equation followed by listing the givens below the respective formulas… That is…

                         Na                      +            H₂O       => NaOH    +         H₂,

Given:      1.76 x 10²⁴ atoms                excess             ---------          ? grams

Convert atoms Na to moles = 1.76 x 10²⁴atoms/6.023 x 10²³atoms/mole

=2.922moles Na produces=>2.922moles H₂(because moles Na=moles H₂).

Convert moles to grams =>2.922moles H₂  x  2.000 grams H₂/mole H₂

=5.8443 grams H₂  

2nd problem, 3.5 moles Na will react with H₂O (in excess) to produce (?) moles of NaOH.

Again write equation and assign values to each formula unit in the equation.

                         Na                      +            H₂O        =>           NaOH    +    H₂,

Given:            3.5moles                       excess                      ? grams       ----

Since coefficients of balanced std equation are equal then moles Na equals moles of NaOH, that is, 3.5 moles Na produces => 3.5 moles NaOH

Convert moles NaOH to grams => 3.5 moles NaOH x 40 g NaOH/mole NaOH =  140 grams NaOH    

3rd problem, 2.75 x 10²⁵ molecules H₂O will react with (?) atoms of Na.

Same procedure, convert to moles, solve problem by ratios then convert to needed dimension at end of problem.

                         Na          +            H₂O                              =>       NaOH    +    H₂

Given:           ? atoms          2.75 x 10²⁵ molecules H₂O    =>     NaOH  + H₂  

Convert to moles =>  2.75 x 10²⁵ molecules H₂O / 6.023 x 10²³ molecules H₂O/mole H₂O = 45.658 moles H₂O =>  45.658 moles Na (equal coefficients)

Convert moles Na to atoms Na  =>   45.658 moles Na x 6.023 x 10²³atoms Na/mole Na = 2.75 x 10²⁵ atoms Na.

Note => Problem 3 could have been solved by inspection b/c coefficients are equal, however, always go through a process that you can justify and defend even if it does take longer. Never assume anything. Depend on what you know, not what you 'think' you know.  

Master the mole concept and you master a lot of chemistry! Good luck.

                             

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