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Calculations Involving Solutions in Reactions

Pearson
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The concentration of a solution is the amount of solute dissolved in a certain amount of solution. There are many different concentration units, including mass percent, molarity, and parts per million, to name a few. The amount of a solute may be expressed in units of grams, milliliters, or moles, depending on the concentration unit. The amount of a solution may be expressed in units of grams, milliliters, or liters, again, depending on the concentration unit. When chemists work with solutions, they often use molarity, a concentration unit that states the number of moles of solute that are dissolved in 1 liter of solution. For example, a 6.0 M HCl solution is 6.0 moles of HCl dissolved in 1 L of solution. When we do calculations using this concentration, we can write two different conversion factors for this expression. Choose a conversion factor for the concentration 34 M Ba(OH)² : Is it a, b, c or d? The correct answer is (b). Capital M is the unit for molarity, which is defined as the moles of solute in one liter of solution. These conversion factors can be used to calculate moles of solute, volume of solution, or mass of solute, when molar mass is calculated. What volume of a 6.0 M HCl solution is needed to obtain 2.0 moles of HCl? Is it a, b, c or d? The correct answer is (b). 2.0 moles of HCl is multiplied by the conversion factor from the molarity. The answer is 0.33 L of HCl solution. When chemical reactions involve aqueous solutions, we use their concentrations along with the balanced chemical equation to determine the concentrations or quantities of other reactants or products. Shown is a pathway going from the volume of one solution to the volume of another solution in a balanced chemical reaction. The molarities of the solutions and the mole-mole factor are used in the calculation. Now let’s use molarity as a conversion factor and mole-mole factors from a balanced chemical equation to determine the volume of a solution. The problem asks how many milliliters of a 0.450 M HCl solution are needed to react with 0.0475 L of a 0.260 M Na₂CO₃ solution? In the reaction, 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid to give 2 moles of sodium chloride, 1 mole of water, and 1 mole of carbon dioxide. Step 1 is to state the given and needed quantities. In this problem, we are given 0.0475 L of a 0.260 M Na₂CO₃ solution. We need milliliters of a 0.450 M HCl solution, and our connection between the two is the mole-mole factor from the balanced chemical equation. Step 2 is to write a plan to calculate the needed quantity. The volume of sodium carbonate (Na₂CO₃) solution is multiplied by the conversion factor from the molarity to give the moles of Na₂CO₃. Then use the mole-mole factor from the balanced chemical equation to give the moles of HCl. Multiply this by the conversion factor from the molarity to give the volume of HCl solution. Use the metric factor to give the milliliters of HCl solution. Step 3 is to write equalities and conversion factors including mole–mole and concentration factors. Shown in yellow, are the equality and the two possible conversion factors for 0.260 M Na₂CO₃. Shown in orange, are the equality and the two possible mole-mole factors from the balanced chemical equation where 1 mole of Na₂CO₃ = 2 moles of HCl. Shown in pink, are the equality and the two possible conversion factors for 0.450 M HCl. Shown in blue, are the equality and the two possible conversion factors for the metric relationship of 1 L equals 1000 mL. Step 4 is to set up the problem to calculate the needed quantity. Our goal is the volume of HCl. The given volume of Na₂CO₃ solution is multiplied by the conversion factor from the molarity, followed by the mole-mole factor from the balanced equation, then the conversion factor from the molarity of the HCl solution, and, finally, the metric factor to convert the liters of HCl solution to milliliters of HCl solution. The answer is 54.9 mL of HCl solution. To review, when you are determining the volume of a reactant in solution from a chemical reaction, you multiply the given volume of the solution by the molarity conversion factor to get moles of A. Use the mole-mole factor from the balanced equation to convert to moles of B. Use the conversion factor from the molarity of B to give the volume of B. Finally, use a metric conversion factor if needed to get required units. How many milliliters of a 0.250 M AgNO₃ solution are needed to react with 0.125 L of a 0.150 M NaCl solution? The balanced chemical equation is 1 mole of aqueous sodium chloride reacts with 1 mole of aqueous silver nitrate to give 1 mole of aqueous sodium nitrate and 1 mole of solid silver chloride. Is it a, b, c or d? The correct answer is (b).
The concentration of a solution is the amount of solute dissolved in a certain amount of solution. There are many different concentration units, including mass percent, molarity, and parts per million, to name a few. The amount of a solute may be expressed in units of grams, milliliters, or moles, depending on the concentration unit. The amount of a solution may be expressed in units of grams, milliliters, or liters, again, depending on the concentration unit. When chemists work with solutions, they often use molarity, a concentration unit that states the number of moles of solute that are dissolved in 1 liter of solution. For example, a 6.0 M HCl solution is 6.0 moles of HCl dissolved in 1 L of solution. When we do calculations using this concentration, we can write two different conversion factors for this expression. Choose a conversion factor for the concentration 34 M Ba(OH)² : Is it a, b, c or d? The correct answer is (b). Capital M is the unit for molarity, which is defined as the moles of solute in one liter of solution. These conversion factors can be used to calculate moles of solute, volume of solution, or mass of solute, when molar mass is calculated. What volume of a 6.0 M HCl solution is needed to obtain 2.0 moles of HCl? Is it a, b, c or d? The correct answer is (b). 2.0 moles of HCl is multiplied by the conversion factor from the molarity. The answer is 0.33 L of HCl solution. When chemical reactions involve aqueous solutions, we use their concentrations along with the balanced chemical equation to determine the concentrations or quantities of other reactants or products. Shown is a pathway going from the volume of one solution to the volume of another solution in a balanced chemical reaction. The molarities of the solutions and the mole-mole factor are used in the calculation. Now let’s use molarity as a conversion factor and mole-mole factors from a balanced chemical equation to determine the volume of a solution. The problem asks how many milliliters of a 0.450 M HCl solution are needed to react with 0.0475 L of a 0.260 M Na₂CO₃ solution? In the reaction, 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid to give 2 moles of sodium chloride, 1 mole of water, and 1 mole of carbon dioxide. Step 1 is to state the given and needed quantities. In this problem, we are given 0.0475 L of a 0.260 M Na₂CO₃ solution. We need milliliters of a 0.450 M HCl solution, and our connection between the two is the mole-mole factor from the balanced chemical equation. Step 2 is to write a plan to calculate the needed quantity. The volume of sodium carbonate (Na₂CO₃) solution is multiplied by the conversion factor from the molarity to give the moles of Na₂CO₃. Then use the mole-mole factor from the balanced chemical equation to give the moles of HCl. Multiply this by the conversion factor from the molarity to give the volume of HCl solution. Use the metric factor to give the milliliters of HCl solution. Step 3 is to write equalities and conversion factors including mole–mole and concentration factors. Shown in yellow, are the equality and the two possible conversion factors for 0.260 M Na₂CO₃. Shown in orange, are the equality and the two possible mole-mole factors from the balanced chemical equation where 1 mole of Na₂CO₃ = 2 moles of HCl. Shown in pink, are the equality and the two possible conversion factors for 0.450 M HCl. Shown in blue, are the equality and the two possible conversion factors for the metric relationship of 1 L equals 1000 mL. Step 4 is to set up the problem to calculate the needed quantity. Our goal is the volume of HCl. The given volume of Na₂CO₃ solution is multiplied by the conversion factor from the molarity, followed by the mole-mole factor from the balanced equation, then the conversion factor from the molarity of the HCl solution, and, finally, the metric factor to convert the liters of HCl solution to milliliters of HCl solution. The answer is 54.9 mL of HCl solution. To review, when you are determining the volume of a reactant in solution from a chemical reaction, you multiply the given volume of the solution by the molarity conversion factor to get moles of A. Use the mole-mole factor from the balanced equation to convert to moles of B. Use the conversion factor from the molarity of B to give the volume of B. Finally, use a metric conversion factor if needed to get required units. How many milliliters of a 0.250 M AgNO₃ solution are needed to react with 0.125 L of a 0.150 M NaCl solution? The balanced chemical equation is 1 mole of aqueous sodium chloride reacts with 1 mole of aqueous silver nitrate to give 1 mole of aqueous sodium nitrate and 1 mole of solid silver chloride. Is it a, b, c or d? The correct answer is (b).