Moles, mass and gas calculations

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Lessons

Using mass and gas volume in stoichiometry calculations
Recap stoichiometry basics
Calculating moles
Molar volume of gas and Avogadro's law.
Worked example: using molar volume and unit conversions

Examples

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Lessons

Calculate the masses and volumes of reactants and products used in chemical reactions.
Consider the reaction:

2C $_6$ H $_{14\;(l)} +$ 19O $_{2\;(g)} \enspace$ → $\enspace$ 12CO $_{2\;(g)} +$ 14H $_2$ O $_{\;(g)}$
1. If 75g of C $_6$ H $_{14}$ is burned, what mass of CO $_2$ would get produced from this reaction?
2. If 240g of H $_2$ O is produced from this reaction, how many moles of C $_6$ H $_{14}$ would be required?
3. At STP, what volume of O $_2$ would be required to produce 85 L of CO $_2$ in this process?
4. What mass of C $_6$ H $_{14}$ would be required to produce 15 moles of CO $_2$ ?
Calculate the masses and volumes of reactants and products used in chemical reactions.
Consider the reaction:

P $_{4\;(s)} +$ 5O $_{2\;(g)} \enspace$ → $\enspace$ P $_4$ O $_{10\;(s)}$
1. At STP, what volume of O $_2$ gas is needed to completely combust 2kg of P $_4$ ?
2. If 25 L of O $_2$ were available, how much mass of P $_4$ could be reacted with?
3. What mass of P $_4$ O $_{10}$ would be produced by this?
Calculate the volume and number of moles of reactants involved in chemical reactions.
Consider the reaction:

2NH $_{3\;(aq)} \,+ \,$ NaOCl $_{\;(aq)} \enspace$ → $\enspace$ N $_2$ H $_{4\;(aq)} \,+ \,$ NaCl $_{\;(aq)} \,+ \,$ H $_2$ O $_{\;(l)}$
1. If 5000 kg of hydrazine (N $_2$ H $_{4}$ ) is required from this industrial process, how much ammonia gas (in L completely dissolved in solution at STP) is required as starting material?
2. How many moles of hydrazine could be produced if only 5 kg of NaOCl was available?
Calculate the volume of reactants required in a chemical process.
Consider the reaction:

SiCl $_{4\;(g)} \, + \,$ 2H $_{2\;(g)} \enspace$ → $\enspace$ Si $_{\;(s)} \, + \,$ 4HCl $_{\;(g)}$

500 mg of Si is required from this process. What is the total volume, in L, of H $_2$ $_{2}$ and SiCl $_4$ $_{4}$ required to produce this?

Topic Notes

In this lesson, we will learn:

To recognize the format of stoichiometry test questions and calculations.
To recall the molar volume of gas at standard temperature and pressure and its meaning.
Methods to calculate number of moles of chemicals in reactions using mass, moles and volume of gas.
To apply Avogadro’s law in finding quantities of gas reactant.

Notes:

We saw in Stoichiometry 1: Introduction that because atoms are so small, they cannot be physically counted when doing reactions in a regular laboratory. We use mass, in grams, as a measure of the amount of substance we have because this is related to moles, the actual number of atoms or molecules present by molar mass. Every substance has a molar mass, measured in grams per mole which is the mass of one mole (or 6.023*10²³ atoms) of that pure substance.

For example, one mole of ¹²C carbon atoms weighs 12g. If you place pure carbon on a weighing balance and it reads as 12g, you have one mole of carbon atoms.

number of moles to relate amounts of substance in reactions

For example, see the reaction:

CH_{4 (g)} + 2O_{2 (g)}

\,

→

\,

CO_{2 (g)} + 2H₂O_(g)

₄

₂

₄

^-1

₂

₄

can measure

The formula: $n(mol) =$ $\large \frac{mass(g)}{M_R (\frac{g}{mol})}$ can be used to calculate the number of moles of a substance when you know the mass, and you can use the periodic table to find atomic or molecular mass M_R of that substance. This applies to elements and molecules, for which you can just add up the elemental mass (e.g. NH₃ = 14+1+1+1=17).
With the moles formula above, you can remember it using the units too, like you would in general algebra: $\large\frac{g}{\frac{g}{mol}}$ = $mol$ and the g will cancel out. You are then left with $\large\frac{1}{\frac{1}{mol}}$ = $\, mol$ which cancels for the mol units.

While we can use mass to find number of moles, we need another relation for gases because it is often quite impractical to measure the mass of a gas. Gases are much easier to measure by their volume, or the amount of 3d space they occupy.

The molar volume of gas at STP, standard temperature and pressure (0°C or 273K, 100 kPa pressure) is 22.4 litres per mole (22.4 L/mol). In other words, one mole of atoms of a pure ideal gas at 0°C will fill 22.4 litres of space.
The molar volume of gas at room temperature (25°C, 298K) and pressure is 24 litres per mole (24 L/mol).

ALWAYS check the conditions of the reaction

Stoichiometry calculations involve unit conversions from one quantity given in the question to an unknown quantity:

To get to moles, use the equation and the molar ratios (the stoichiometric ratio) shown.
To get to volume, use the molar volume of gas constants.
To get to mass, use the atomic/molecular masses shown in the periodic table.

Introduction: Unit conversions in chemistry

O_{2 (g)} + 2Mg_(s)

\,

→

\,

2MgO_(s)

standard temperature

72.9

g \, Mg =

\large \frac{1 \, mol \, Mg} {24.3 \, g \, Mg} * \frac{1 \, mol \, O_{2}} {2 \, mol \, Mg} * \frac{22.4 \, L \, O_{2}} {1 \, mol \, O_{2}}

=

33.6 L O₂

As mentioned above, the molar volumes of gas at RTP and STP are based on Avogadro’s law which states that at constant temperature and pressure, equal volumes of any two gases have the same number of moles.
This can be expressed as:

\large \frac{V_{1}} {n_{1}} = \frac{V_{2}} {n_{2}}

Worked example: Using Avogadro’s law At STP, 10 L of O₂ gas contains 0.45 mol of O₂ molecules. If 3.5 moles more O₂ gas is added to the container, what is the new volume?

₂

V_{2} =

\large \frac{V_{1}} {n_{1}}

\, * \, n_{2}

₁

₂

V_{2} =

\large \frac{10} {0.45} \,* \,

3.95 = 87.8

L \,O_{2}

Moles, mass and gas calculations

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Moles, mass and gas calculations

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