CAN you please solve this? please please please

115.2ml of water should be added.

Explanation:

We start with 46.2ml of 1.396M sugar solution;

We want to end up with a 0.56M sugar solution;

Finding the number of moles in the sugar solution to begin with:

moles of sugar in solution initially = 46.2 × 1.396 = 64.4952

Vd = volume of desired solution (i.e. 0.56M solution)

w = volume of water added

Vd = (46.2 + w)

Using the mol = vol' × conc'n equation:

64.4952 = Vd × 0.56

Vd = 64.4952/0.56

Vd = 115.17

46.2 + w = 115.17

w = 115.17 - 46.2

w = 68.97ml

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