How many moles do you need to prepare 1.675 L of a 8.50 M NaOH solution? Round your answer to 2 decimal places.

0.20 mol's

Explanation:

1.675 L = 1.675 dm^3

moles = V/(conc):

moles = 1.675/(8.5)

moles = 0.1970... --> 0.20

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