What volume if dilute hno3 solution (0.150 mol/l) is required to obtain 10.0 grams pure hno3

1.058 L

Explanation:

Data given:

Molarity of HNO₃ solution = 0.150 mol/L

Mass of HNO₃ = 10 g

volume of HNO₃ solution = ?

Solution:

First we have to know the number of moles for 10 g of HNO₃

Now calculate mass to moles

Formula used

                no. of moles = Mass of HNO₃ / Molar mass . . . . . . . . . . (1)

molar mass of HNO₃ = 1 + 14 + 3(16)

molar mass of NaNO₃ = 63 g/mol

Put value in equation 1

              no. of moles =  10 g / 63 g/mol

              no. of moles =  0.1587 mol

Now to find volume

Formula used for Molarity

                Molarity = number of moles of solute / volume of solution

Rearrange above equation

   volume of solution = number of moles of solute / Molarity . . . . . . . (2)

Put values in equation 2

         volume of solution = 0.1587 mol / 0.150 mol/L

         volume of solution =  1.058 L

volume of dilute HNO₃ solution =  1.058 L

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