Question

How many milliliters of \(0.258 \mathrm{M} \mathrm{NiCl}_{2}\) solution are needed to react completely with \(20.0 \mathrm{~mL}\) of \(0.153 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution? How many grams of \(\mathrm{NiCO}_{3}\) will be formed? The reaction is \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NiCl}_{2}(a q) \longrightarrow \mathrm{NiCO}_{3}(s)+2 \mathrm{NaCl}(a q)\)

Short answer

You will need 11.9 mL of 0.258 M NiCl2 solution to react with 20.0 mL of 0.153 M Na2CO3 solution, resulting in the formation of 0.3634 grams of NiCO3.

Step-by-step solution

Write the Balanced Chemical Equation

Ensure that the chemical equation for the reaction is balanced. The provided equation is already balanced and is as follows: \[\mathrm{Na}_{2} \mathrm{CO}_{3}(aq)+\mathrm{NiCl}_{2}(aq) \longrightarrow \mathrm{NiCO}_{3}(s)+2 \mathrm{NaCl}(aq)\]

Calculate Moles of Na2CO3

Use the concentration and volume to calculate the moles of Na2CO3. The molar concentration (M) is moles per liter (mol/L), so multiply the volume in liters by the concentration in molarity: \[\text{Moles of } \mathrm{Na}_{2} \mathrm{CO}_{3} = 0.153 \mathrm{M} \times 0.0200 \mathrm{L} = 0.00306 \mathrm{mol}\]

Determine Moles of NiCl2 Required

From the balanced equation, the stoichiometry is 1:1 for Na2CO3 to NiCl2. Use the moles of Na2CO3 to find the moles of NiCl2 needed: \[\text{Moles of } \mathrm{NiCl}_{2} = \text{Moles of } \mathrm{Na}_{2} \mathrm{CO}_{3} = 0.00306 \mathrm{mol}\]

Calculate Volume of NiCl2 Solution Needed

Divide the moles of NiCl2 by its molarity to find the volume in liters, then convert to milliliters: \[\text{Volume of } \mathrm{NiCl}_{2} = \frac{0.00306 \mathrm{mol}}{0.258 \mathrm{M}} = 0.0119 \mathrm{L}\] \[0.0119 \mathrm{L} \times 1000 \mathrm{mL/L} = 11.9 \mathrm{mL}\]

Determine Mass of NiCO3 Formed

Calculate the molar mass of NiCO3 and then use the moles of NiCO3 (same as moles of NiCl2 and Na2CO3) to find the mass: Molar mass of NiCO3 = 58.69 (Ni) + 12.01 (C) + 3\times16.00 (O) = 118.69 g/mol \[\text{Mass of } \mathrm{NiCO}_{3} = 0.00306 \mathrm{mol} \times 118.69 \mathrm{g/mol} = 0.3634 \mathrm{g}\]

Key concepts

Balancing Chemical Equations

Balancing chemical equations is essential in chemistry to ensure the law of conservation of mass is adhered to, meaning the same number of atoms of each element must be present on both sides of the equation. In the given exercise, the balanced equation \[\mathrm{Na}_2 \mathrm{CO}_3(aq) + \mathrm{NiCl}_2(aq) \longrightarrow \mathrm{NiCO}_3(s) + 2 \mathrm{NaCl}(aq)\]shows a stoichiometric ratio of 1:1 between \(\mathrm{Na}_2 \mathrm{CO}_3\) and \(\mathrm{NiCl}_2\), highlighting the direct relationship in the amounts of reactants and products. When attempting exercises involving chemical reactions, always ensure to balance the equation first. It provides the ratio of reactants to products, which is critical for stoichiometric calculations.

Molarity and Concentration

Molarity is a measure of concentration, specifically the number of moles of a solute per liter of solution, expressed as moles per liter (M). In the context of the exercise, molarity helps us quantify reactants in liquid solutions. For instance,\[0.153 \mathrm{M} \times 0.0200 \mathrm{L} = 0.00306 \mathrm{mol}\]measures the amount of \(\mathrm{Na}_2 \mathrm{CO}_3\) in moles. Understanding molarity is vital for conducting stoichiometric calculations as it directly relates to the quantities of substances involved.
To grasp these concepts, always remember that molarity describes the 'strength' of the solution in terms of solute amount, while the volume of the solution tells us 'how much' of it we have.

Mole Concept

The mole concept is central to stoichiometry, providing a link between the microscopic world of atoms and the macroscopic world we measure in the lab. One mole is defined as the amount of substance containing as many entities (such as atoms or molecules) as there are atoms in 12 grams of carbon-12. It offers a way to convert between the mass of substance and the number of particles. In our example, when we calculate the mass of \(\mathrm{NiCO}_3\) formed, we use the molar mass and the moles of the compound:\[0.00306 \mathrm{mol} \times 118.69 \mathrm{g/mol} = 0.3634 \mathrm{g}\]This step is crucial as it makes abstract quantities like 'moles' tangible by converting them to measurable grams.

Stoichiometric Calculations

Stoichiometric calculations involve using the relationships indicated by a balanced chemical equation to determine the quantities of reactants and products. These calculations can include finding the amount of reactants needed to produce a set amount of product or the quantity of product formed from certain reactants. In the provided exercise, stoichiometry is used to find both the volume of \(\mathrm{NiCl}_2\) solution required and the mass of \(\mathrm{NiCO}_3\) precipitate formed from the reaction with \(\mathrm{Na}_2 \mathrm{CO}_3\). The steps of the solution include calculating moles from volume and molarity, using the stoichiometry to relate moles of different chemicals, and finally converting moles back to grams or milliliters as needed. Grasping stoichiometric calculations is fundamental for predicting the outcomes of chemical reactions in quantitative terms.