Question

the molarity of NO3 in the solution after 2litres of 3M AgNo3 is mixed with 3 litres 1M Bacl2​

Answer 1

plz reply if your doubt isn't solve

Answer 2

The molarity of $NO_3^-$ in the solution after 2 litres of 3M  $AgNO_3$ is mixed with 3 litres 1M $BaCl_2$

Explanation:

To calculate the moles of cadmium nitrate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$    

1. Molarity of silver nitrate = 3 M

Volume of silver nitrate = 2 L  

$3M=\frac{\text{Moles of silver nitrate}}{2L}\\\\\text{Moles of silver nitrate}=6mol$

2. Molarity of barium chloride= 3 M

Volume of barium chloride = 1 L  

$3M=\frac{\text{Moles of barium chloride}}{1L}\\\\\text{Moles of barium chloride}=3mol$

The chemical equation for the reaction of silver nitrate and barioum chloride follows:

$2AgNO_3+BaCl_2\rightarrow Ba(NO_3)_2+2AgCl$

2 moles of  $AgNO_3$ reacts with 1 mole of $BaCl_2$

6 moles of  $AgNO_3$ reacts with =$\frac{1}{2}\times 6=3$ moles of $BaCl_2$

2 moles of  $AgNO_3$ produces = 1 mole of $Ba(NO_3)_2$

6 moles of  $AgNO_3$ produces =$\frac{1}{2}\times 6=3$ moles of $Ba(NO_3)_2$

moles of $(NO_3)^-=3\times 2= 6$

Molarity of $NO_3^-$ = $\frac{moles}{volume}=\frac{6moles}{5L}=1.2M$

Learn more about molarity

https://brainly.com/question/13199832

https://brainly.com/question/10608366