How many milliliters of a 0.250 M BaCl2 solution are needed to react with 0.0325 L of a 0.160 M Na2SO4 solution? Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)

Question

How many milliliters of a 0.250 M [latex]BaCl_{2}[/latex] solution are needed to react with 0.0325 L of a 0.160 M [latex]Na_{2}SO_{4}[/latex] solution?

[latex]Na_{2}SO_{4}(aq) + BaCl_{2}(aq) → BaSO_{4}(s) + 2NaCl(aq)[/latex]

Accepted Answer

STEP 1 State the given and needed quantities.

ANALYZE THE PROBLEM	Given	Need	Connect
	0.0325 L of 0.160 M [latex]Na_{2}SO_{4}[/latex] solution, 0.250 M [latex]BaCl_{2}[/latex] solution	milliliters of [latex]BaCl_{2}[/latex] solution	mole-mole factor
	Equation
	[latex]Na_{2}SO_{4}(aq) + BaCl_{2}(aq) → BaSO_{4}(s) + 2NaCl(aq)[/latex]

STEP 2 Write a plan to calculate the needed quantity.

[latex] liters of Na_{2}SO_{4} solution \boxed{Molarity} → moles of Na_{2}SO_{4} \boxed{\begin{array}{l}mole-mole \ factor\end{array}} → moles of BaCl_{2} \boxed{Molarity} → liters of BaCl_{2} solution \boxed{\begin{array}{l}Metric\factor\end{array}}→ milliliters of BaCl_{2} solution[/latex]

STEP 3 Write equalities and conversion factors including mole-mole and concentration factors.

[latex] \begin{array}{r c}\boxed{\begin{matrix} 1 L of solution = 0.160 mole of Na_{2}SO_{4} \\frac{0.160 mole Na_{2}SO_{4}}{1 L solution } ext{ and } \frac{1 L solution}{0.160 mole Na_{2}SO_{4}} \end{matrix}} & \boxed{\begin{matrix} 1 mole of Na_{2}SO_{4}= 1 mole of BaCl_{2} \ \frac{1 mole Na_{2}SO_{4}}{ 1 mole BaCl_{2}} ext{ and }\frac{1 mole BaCl_{2}}{1 mole Na_{2}SO_{4}} \end{matrix}}\end{array} [/latex]

[latex] \begin{array}{r c}\boxed{\begin{matrix} 1 L of solution = 0.250 mole of BaCl_{2}\\frac{0.250 mole BaCl_{2}}{1 L solution } ext{ and } \frac{1 L solution}{0.250 mole BaCl_{2}} \end{matrix}} & \boxed{\begin{matrix}1 L = 1000 mL \ \frac{1000 mL}{ 1 L } ext{ and }\frac{1 L }{1000 mL} \end{matrix}}\end{array} [/latex]

STEP 4 Set up the problem to calculate the needed quantity.

[latex] 0.0325 \cancel{L solution} imes \boxed{\frac{0.160 \cancel{mole Na_{2}SO_{4}}}{1 \cancel{L solution}} } imes \boxed{\frac{1 \cancel{mole BaCl_{2}}}{1 \cancel{ mole Na_{2}SO_{4}}} } imes \boxed{\frac{1 \cancel{L solution}}{0.250 \cancel{ mole BaCl_{2}}} } \boxed{\frac{1000 mL BaCl_{2} solution }{1 \cancel{L solution}} }[/latex]

[latex]= 20.8 mL of BaCl_{2} solution [/latex]

Question 9.13: How many milliliters of a 0.250 M BaCl2 solution are needed ...

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