How many atoms are present in 1.4 grams of aluminum?

3.2x10^23

Explanation:

Start by finding the molar mass of aluminum.

Then turn  grams into moles by dividing grams of substance by molar mass. 1.4g÷26.98g

Then multiply that by 6.022x10^23 atoms. 0.0519x6.022x10^23

You can use dimensional analysis to solve this type of problem.

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