A student sets up and solves the following equation to solve a problem in solution stoichiometry. Fill in the missing part of the student's equation.
[tex](0.20 L)\left(\frac{1 mL}{10^{-3} L}\right)(0.60 \square)\left(\frac{1 kg}{10^3 g}\right)=0.12 kg[/tex]
Answer (1)
Convert liters to milliliters: 0.20 L = 200 m L .
Convert kilograms to grams: 0.12 k g = 120 g .
Isolate the unknown term x in the equation: ( 200 m L ) ( 0.60 x ) = 120 g .
Solve for x : x = 120 m L 120 g = 1 m L g .
The missing part of the equation is 1 m L g .
Explanation
Understanding the Problem We are given the equation (0.20 L)\\\left(\\frac{1 mL}{10^{-3} L}\\right)(0.60 \\square)\\\\left(\\frac{1 kg}{10^3 g}\\right)=0.12 kg and we need to find the missing term represented by the square. Let's denote the missing term by x and its units. The equation can be rewritten as: ( 0.20 L ) l e f t ( f r a c 1 m L 1 0 − 3 L r i g h t ) ( 0.60 x ) l e f t ( f r a c 1 k g 1 0 3 g r i g h t ) = 0.12 k g .
Converting Units First, let's convert liters (L) to milliliters (mL) using the conversion factor f r a c 1 m L 1 0 − 3 L . We have 0.20 L = 0.20 L t im es f r a c 1 m L 1 0 − 3 L = 200 m L . So the equation becomes: ( 200 m L ) ( 0.60 x ) l e f t ( f r a c 1 k g 1 0 3 g r i g h t ) = 0.12 k g .
Converting Units Next, let's convert kilograms (kg) to grams (g) using the conversion factor f r a c 1 0 3 g 1 k g . We have 0.12 k g = 0.12 k g t im es f r a c 1 0 3 g 1 k g = 120 g . Now the equation is: ( 200 m L ) ( 0.60 x ) l e f t ( f r a c 1 k g 1 0 3 g r i g h t ) = 120 g .
Isolating the Unknown Now, let's isolate x . We can rewrite the equation as: ( 200 m L ) ( 0.60 x ) = 120 g t im es f r a c 1 0 3 g 1 k g = 120 t im es 1 0 3 f r a c g 2 k g . However, we want to find x such that ( 200 m L ) ( 0.60 x ) l e f t ( f r a c 1 k g 1 0 3 g r i g h t ) = 120 g . So, we have ( 200 m L ) ( 0.60 x ) = 120 g t im es f r a c 1 0 3 g 1 k g = 120 g .
Solving for x Now we can solve for x : x = f r a c 120 g ( 200 m L ) ( 0.60 ) = f r a c 120 g 120 m L = 1 f r a c g m L . Therefore, the missing term is 1 f r a c g m L . This represents the density of the solution.
Final Equation Thus, the complete equation is ( 0.20 L ) l e f t ( f r a c 1 m L 1 0 − 3 L r i g h t ) ( 0.60 f r a c g m L ) l e f t ( f r a c 1 k g 1 0 3 g r i g h t ) = 0.12 k g .
Examples
In chemistry, solution stoichiometry is used daily to calculate the amount of reactants needed or products formed in a chemical reaction. For example, if you are titrating an acid with a base, you need to know the concentration and volume of each solution to determine the equivalence point. The equation we solved is similar to calculating the mass of a solute in a solution given its volume and concentration. This is crucial in many laboratory experiments and industrial processes where precise measurements are essential for successful outcomes. For instance, a pharmaceutical company needs to accurately determine the amount of active ingredient in a drug formulation to ensure its safety and efficacy. This involves calculations similar to the one we just performed, where we convert between volumes, concentrations, and masses using appropriate conversion factors.
