Answer:
0.019 L or 19.1 mL
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
Answer:
19 mL of CaBr₂ solution should be used to react with AgNO₃
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react:
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?
[tex]moles of AgNO_{3}=\frac{0.05L*0.115 moles}{1 L}[/tex]
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 ?
[tex]moles of CaBr_{2} =\frac{5.75*10^{-3}moles of AgNO_{3}*1 mole of CaBr_{2} }{2moles of AgNO_{3}}[/tex]
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?
[tex]volume=\frac{2.875*10^{-3} moles*1L}{0.150 moles}[/tex]
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
