A student has two solutions of a substance;Solution-1:25M , 400ml and Solution-2:30M , 300M. What is the molarity of the final solutions if these two solutions are mixed?

The molarity of the final solution  is 1.7 M

Explanation:

The parameters given are;

First solution = 400 ml of 1.25 M

Second solution = 300 ml of 2.30 M

Therefore, we have;

First solution contains 400/1000 * 1.25 moles = 0.5 moles of the substance

Second solution contains 300/1000 * 2.30 moles = 0.69 moles of the

Hence the sum of the two solutions contains 0.5 + 0.69 = 1.19 moles of the substance

The volume of the sum of the two solutions = 400 ml + 300 ml = 700 ml

Hence we have the concentration of the final solution presented as follows;

700 ml contains 1.19 moles of the substance

Therefore;

1000 ml will contain 1000/700 * 1.19 = 1.7 moles

The molarity of the final solution = The number of moles per 1000 ml = 1.7 M.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2