Question

What volume of \(1.000 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) will react with \(342 \mathrm{~mL}\) of \(0.733 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} ?\) \(3 \mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow 2 \mathrm{Na}_{3} \mathrm{PO}_{4}+3 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{CO}_{2}\)

Short answer

375.6 mL of \(\mathrm{Na}_2\mathrm{CO}_3\) is needed.

Step-by-step solution

Write the Balanced Chemical Equation

The chemical reaction provided is already balanced: \(3\, \mathrm{Na}_2\mathrm{CO}_3 + 2\, \mathrm{H}_3\mathrm{PO}_4 \rightarrow 2 \mathrm{Na}_3\mathrm{PO}_4 + 3 \mathrm{H}_2\mathrm{O} + 3\, \mathrm{CO}_2\). This shows that 3 moles of \(\mathrm{Na}_2\mathrm{CO}_3\) react with 2 moles of \(\mathrm{H}_3\mathrm{PO}_4\).

Convert Volume of \(\mathrm{H}_3\mathrm{PO}_4\) to Moles

Use the molarity formula, \( \text{Moles} = \text{Molarity} \times \text{Volume (L)} \). First, convert 342 mL to liters: \(\frac{342}{1000} = 0.342\, \text{L}\). Then, calculate the moles of \(\mathrm{H}_3\mathrm{PO}_4\): \(0.733\, \text{mol/L} \times 0.342\, \text{L} = 0.250386\, \text{mol}\).

Determine Moles of \(\mathrm{Na}_2\mathrm{CO}_3\) Needed

According to the balanced equation, \(\frac{3}{2}\) moles of \(\mathrm{Na}_2\mathrm{CO}_3\) are needed for each mole of \(\mathrm{H}_3\mathrm{PO}_4\). Thus, \(0.250386\, \text{mol} \times \frac{3}{2} = 0.375579\, \text{mol} \) of \(\mathrm{Na}_2\mathrm{CO}_3\) is required.

Calculate the Volume of \(\mathrm{Na}_2\mathrm{CO}_3\) Solution

Use the molarity formula rearranged for volume: \( \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}}\). With the moles of \(\mathrm{Na}_2\mathrm{CO}_3\) calculated as 0.375579 and its molarity as 1.000 M, the volume required is \(\frac{0.375579\, \text{mol}}{1.000\, \text{mol/L}} = 0.375579\, \text{L}\). Convert this to mL: \(0.375579\, \text{L} \times 1000 = 375.579\, \text{mL}\).

Key concepts

Balanced Chemical Equation

Understanding balanced chemical equations is crucial in stoichiometry as it provides a clear roadmap of how substances interact in a chemical reaction. A balanced chemical equation has equal numbers of each type of atom on both sides of the equation, ensuring that matter is conserved as per the law of conservation of mass. In a chemical equation, the coefficients before each compound indicate the number of moles required or produced.
For example, in the equation:
\[ 3 \, \mathrm{Na}_2\mathrm{CO}_3 + 2 \, \mathrm{H}_3\mathrm{PO}_4 \rightarrow 2 \mathrm{Na}_3\mathrm{PO}_4 + 3 \, \mathrm{H}_2\mathrm{O} + 3 \, \mathrm{CO}_2 \]
we see that 3 moles of sodium carbonate \( \mathrm{Na}_2\mathrm{CO}_3 \) react with 2 moles of phosphoric acid \( \mathrm{H}_3\mathrm{PO}_4 \). This balanced equation helps us know exactly how much of each substance is required and what is produced.

Molarity

Molarity is the concentration of a solution expressed as the number of moles of solute per liter of solution. It's a key concept in chemistry for preparing and performing reactions with solutions. Molarity is calculated using the formula: \[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \] To explore this with our example for phosphoric acid \( \mathrm{H}_3\mathrm{PO}_4 \), the given molarity is 0.733 M, which means each liter of the solution contains 0.733 moles of \( \mathrm{H}_3\mathrm{PO}_4 \). This information allows us to calculate the number of moles present when given a specific volume, proving invaluable in stoichiometry.
For instance, if we're working with a volume of 0.342 liters, the moles of \( \mathrm{H}_3\mathrm{PO}_4 \) would be calculated by multiplying the molarity by the volume in liters.

Volume Conversion

In stoichiometry and chemistry, converting volumes between different units is a frequent task. The most common conversion, especially in solution chemistry, is between milliliters (mL) and liters (L).
Remember: 1 L = 1000 mL.
To convert milliliters to liters, divide the number of milliliters by 1000. For our phosphoric acid volume, converting 342 mL to liters involves calculating:
\[ \frac{342 \text{ mL}}{1000} = 0.342 \text{ L} \] This conversion is essential before performing any calculations involving molarity, as the standard unit for volume in the molarity equation is liters. Proper unit conversion ensures accuracy in chemical computations and balanced reactions.

Mole Calculation

Calculating moles accurately is foundational in stoichiometry, allowing one to interconvert between quantities of substances in a reaction. Using the relations from balanced equations and molarities allows us to find necessary moles for reactions.
For example, given the moles of \( \mathrm{H}_3\mathrm{PO}_4 \) and knowing its reaction ratio with \( \mathrm{Na}_2\mathrm{CO}_3 \) is \( \frac{3}{2} \), we can determine the moles of \( \mathrm{Na}_2\mathrm{CO}_3 \) required. Multiply the moles of phosphoric acid by \( \frac{3}{2} \):\[ 0.250386 \, \text{mol} \times \frac{3}{2} = 0.375579 \, \text{mol} \] These calculations show the integration of mole concept with balanced equations and molarity, helping ascertain the amount of reactants or products in a given chemical equation.