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Stoichiometric reactions

The masses of different products produced in a reaction are determined by following the steps below.

Step 1: The masses of the reactants are converted to moles using the concept of molar mass.

Step 2: The number of moles of products formed is determined from the ratios of coefficients.

Step 3: The masses of the products are found by converting moles to mass using the molar mass.

$$$\ce {2H2 {(g)} + O2 {(g)} -> 2H2O {(l)}}$$$

Given that 4 g of $$\ce {H2}$$ react, we must first convert this value to moles. 4 g of $$\ce{H2}$$ contains 2 moles of $$\ce{H2}$$ (using molar mass of $$\ce{H2}$$).

Based on the coefficients, 2 moles of $$\ce{H2O}$$ are produced for every 2 moles of $$\ce{H2}$$ that react. The amount of $$\ce {H2O}$$ produced is thus 2 moles.

The 2 moles of $$\ce{H2O}$$ amount to 36 g of $$\ce {H2O}$$.

The volume of a gas is related to the number of moles (and thus the number of molecules) in the gas according to the relationship below.

1 mole of almost any gas occupies a volume of approximately $$24\cdm(=\text{24 L})$$ at room temperature and pressure.

This relationship holds regardless of the chemical identity of the gas.

If the pressure and temperature are constant, then $$\text{1 L}$$ of $$\ce{H2}$$ gas has the same number of moles (and therefore the same number of molecules) as $$\text{1 L}$$ of $$\ce{C4H10}$$.

This relationship is useful in reactions involving gases because gases are typically measured in terms of volume instead of mass.

$$$\ce{Mg {(s)} + 2HCl {(aq)} → MgCl2 {(aq)} + H2 {(g)}}$$$

Given that 2 moles of $$\ce{Mg}$$ react, 2 moles of $$\ce{H2}$$ will be produced.

At room temperature and pressure, 2 moles of $$\ce {H2}$$ has a volume of $$48\cdm.$$

The volume of a liquid can be converted to mass using a relationship called density, which gives the amount of mass per volume.

In reactions involving liquids, density is a useful relationship because liquids are typically measured in terms of volume.

Using density, volume can be converted to mass, which can then be converted to the number of moles using molar mass.

The number of moles produced in the reaction can be converted back to volume using the same relationships.

$$$\ce{2C6H14 {(l)} + 19O2 {(g)} → 12CO2 {(g)} + 14H2O {(g)}}$$$

$$\ce{C6H14}$$ has a density of $$\text{0.65 g/mL}$$.

Given that 100 mL of $$\ce{C6H14}$$ are used, 65 g of $$\ce{C6H14}$$ are present. This is converted into 0.756 mol (correct to 3 significant figures) of $$\ce{C6H14}$$.

2 moles of $$\ce{C6H14}$$ gives 12 moles of $$\ce{CO2}$$. 0.756 mole of $$\ce{C6H14}$$ thus produces 4.54 moles of $$\ce{CO2}$$.

Concentration is a measure of the amount of a substance dissolved in solution.

Concentration is expressed as the number of moles per$$\cdm$$ $$(\text{mol}/\cdm)$$ and sometimes as the mass per$$\cdm$$ $$(\ug/\cdm)$$.

Aqueous substances are often measured in terms of concentration. When solving reaction problems involving aqueous compounds, concentration must be converted to moles by multiplying by the volume.

If concentration is expressed as mass per volume, then mass is found by multiplying by volume. Molar mass must then be used to convert to moles.

$$$\text{Mass (g)} = \text{Concentration (g/}\cdm) \times \text{Volume }(\cdm)$$$ $$$\text{Number of moles (mol)} = \text{Mass (g)} \div \text{Molar mass (g/mol)}$$$

The moles of the reactants are then used to find the moles of products produced.

If the concentration of a $$\ce{H2SO4}$$ solution is $$\text{10 mol/}\cdm$$ and there are $$2\cdm$$, there are 20 moles of $$\ce{H2SO4}$$ in solution.

The limiting reactant is the reactant that is used up first, causing the reaction to stop.

When multiple reactants are present, one of them will probably be used up first. Once the limiting reactant is gone, the reaction cannot proceed. The reactants that are not limiting are left over.

The limiting reactant can be determined if the number of moles of each of the reactants is known.

The initial number of moles of each reactant is compared with the coefficient ratios of each substance to determine which reactant will be used up first.

$$$\ce {NaOH {(aq)} + HCl {(aq)} -> NaCl {(aq)} + H2O {(l)}}$$$

1 mole of $$\ce {NaOH}$$ only requires 1 mole of $$\ce {HCl}$$ for complete reaction.

If 1.5 moles of $$\ce {HCl}$$ are added to 1 mole of $$\ce {NaOH}$$, $$\ce {HCl}$$ is clearly in excess. $$\ce {NaOH}$$ is the limiting reactant.