Chemistry, 10.03.2021 18:00 reedj7736

C3H5N3O9(s) --> N2(g) + CO2(g) + H2O(g) + O2(g) Solid nitroglycerine explodes when heated, producing several different gases (according to the reaction above). If a government scientist explodes 777 g of nitroglycerine, How many grams of each gas should be produced? How many molecules of each gas will be produced?

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Answers

Answer from: Anshuman2002

- 143.6 g of N₂ ; 3.08×10²⁴ molecules of N₂

- 451.4 g of CO₂ ; 6.17×10²⁴ molecules of CO₂

- 153.9 g of H₂O ; 5.14×10²⁴ molecules of H₂O

- 27.3 g of O₂ ; 5.14×10²³ molecules of O₂

Explanation:

This is a reaction of decomposition:

4C₃H₅N₃O₉(s) → 6N₂(g) + 12CO₂(g) + 10H₂O(g) + O₂(g)

4 moles of solid nitroglycerine decompose to 6 moles of nitrogen, 12 moles of carbon dioxide, 10 moles of water vapor and 1 mol of oxygen

We convert mass to moles → 777 g. 1mol/ 227g = 3.42 moles

4 moles of C₃H₅N₃O₉ can decompose to:

6 moles of N₂ 12 moles of CO₂ ___ 10 moles of H₂O ___ 1 mol of O₂

Then, 3.42 moles of C₃H₅N₃O₉ may decompose to:

(3.42 . 6) / 4 = 5.13 moles of N₂

(3.42 . 12) / 4 = 10.26 moles of CO₂

(3.42 . 10) / 4 = 8.55 moles of water vapor

(3.42 . 1) / 4 = 0.855 moles of oxygen

We convert the moles to mass:

5.13 mol . 28 g/mol = 143.6 g of N₂

10.26 mol . 44 g/mol = 451.4 g of CO₂

8.55 mol . 18 g/mol = 153.9 g of H₂O

0.855 mol . 28 g/mol = 27.3 g of O₂

We count the atoms:

5.13 mol . 6.02×10²³ molecules /mol = 3.08×10²⁴ molecules of N₂

10.26 mol . 6.02×10²³ molecules /mol = 6.17×10²⁴ molecules of CO₂

8.55 mol . 6.02×10²³ molecules /mol = 5.14×10²⁴ molecules of H₂O

0.855 mol . 6.02×10²³ molecules /mol = 5.14×10²³ molecules of O₂

Answer from: Quest

why is arctic sea ice important? arctic sea ice keeps the polar regions cool and moderate global climate. sea ice has a bright surface; 80 percent of the sunlight that strikes it is reflected back into space. as sea ice melts in the summer, it exposes the dark ocean surface.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.