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What volume of 1.46 M CrBr2 solution is required to react with 50.0 mL of 0.210 M Na2CO3 solution?

User Aravind NC
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Final answer:

To find the volume of the CrBr2 solution required to react with the Na2CO3 solution, we can use the concept of stoichiometry and molar ratios. By setting up a proportion and solving for the unknown volume, we find that 348.1 mL of the 1.46 M CrBr2 solution is required.

Step-by-step explanation:

To find the volume of the CrBr2 solution required to react with the Na2CO3 solution, we can use the concept of stoichiometry. First, we need to balance the equation:

CrBr2(aq) + Na2CO3(aq) → Cr(CO3)2(aq) + 2NaBr(aq)

From the balanced equation, we can see that one mole of CrBr2 reacts with one mole of Na2CO3. Using the given molarities and volumes, we can set up a proportion:

(1.46 M CrBr2) / (x mL CrBr2) = (0.210 M Na2CO3) / (50.0 mL Na2CO3)

Cross multiplying and solving for x, we get:

x = (1.46 M CrBr2) * (50.0 mL Na2CO3) / (0.210 M Na2CO3)

x = 348.1 mL

User Moj
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