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murzikaleks [220]
4 years ago
9

1. Write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin (C3H5NO33

) into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide.
Chemistry
1 answer:
zvonat [6]4 years ago
6 0

Explanation:

A balance chemical equation is a representation of a chemical reaction in which reactants are written on left hand side followed by arrow pointing in right direction along with product on the right-hand side. A balanced chemical equation follows the law of conservation of mass.

The balanced chemical equation for decomposition of nitroglycerin is given as:

4C_3H_5N_3O_{9}(l)\rightarrow 6N_2(g)+O_2(g)+10H_2O(g)+12CO_2(g)

4 moles of nitroglycerin on decomposition gives 6 mole of dinitrogen, 1 mole of dioxygen, 10 moles of water and 12 moles of carbon dioxide.

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Calculate the volume of an object with dimensions measuring: 15cm x 6cm x 10cm
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4. Magnesium and oxygen undergo a chemical reaction to
erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

3 0
2 years ago
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