Question

Need help please: How many mL of 0.150 Molarity CaBr2 solution must be used to react with 50 mL of 0.115 Molarity AgNO3?

CaBr2 + 2AgNO3 --> Ca(No3)2 + 2AgBr

Answer 1

19 mL of CaBr₂ solution should be used to react with AgNO₃

Step-by-step explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react:

• AgNO₃: 2 moles
• CaBr₂: 1 mole

50 ml of 0.115 M AgNO₃ react. Since molarity is the number of moles of solute per liter of solution, then you can do the following rule of three: if there are 0.115 moles of AgNO₃ in 1 L, in 0.05 L (being 1 L = 1000 mL, then 50 mL = 0.05 L How many moles are there?

`moles of AgNO_(3)=(0.05L*0.115 moles)/(1 L)`

moles of AgNO₃=5.75*10⁻³

Now we can make a new rule of three to calculate the amount of moles necessary for CaBr₂ to react with moles of AgNO₂: if stoichiometry 2 moles of AgNO₃ react with 1 mole of CaBr₂, 5.75*10⁻³ moles of AgNO₃ with how many moles of CaBr₂ react ?

`moles of CaBr_(2) =(5.75*10^(-3)moles of AgNO_(3)*1 mole of CaBr_(2) )/(2moles of AgNO_(3))`

moles of CaBr₂=2.875*10⁻³ moles

Being 0.150 the molarity of the CaBr₂ compound, you can apply the following rule of three: if by definition of molarity 0.150 moles are in 1 L, 2.875 * 10⁻³ moles in how much volume is it present?

`volume=(2.875*10^(-3) moles*1L)/(0.150 moles)`

volume=0.019 L

Being 1 L = 1000 mL, then 0.019 L = 19 mL

19 mL of CaBr₂ solution should be used to react with AgNO₃

Answer 2

0.019 L or 19.1 mL

Step-by-step explanation:

Given the equation of the reaction; we have:

CaBr2 + 2AgNO3 --> Ca(No3)2 + 2AgBr

We must first obtain the number of moles of AgNO3 reacted. This can be obtained from;

Concentration of AgNO3 reacted= 0.115 M

Volume of AgNO3 reacted= 50 ml

Since

Number of moles of AgNO3= concentration of AgNO3 × volume of AgNO3

Number of moles = 0.115 M × 50/1000 = 5.75 ×10^-3 moles

From the reaction equation;

2 moles of AgNO3 reacts with 1 mole of CaBr2

5.75 ×10^-3 moles of AgNO3 reacts with 5.75 ×10^-3 moles × 1 /2 = 2.875×10^-3 moles of CaBr2

Now;

Number of moles of CaBr2= concentration of CaBr2 × volume of CaBr2

Number of moles of CaBr2= 2.875×10^-3 moles

Volume of CaBr2= ????

Concentration of CaBr2= 0.150 M

Hence;

Volume of CaBr2= number of moles of CaBr2/ concentration of CaBr2

Volume of CaBr2= 2.875×10^-3 moles / 0.150

Volume of CaBr2= 0.019 L or 19.1 mL