Aparticular brand of gasoline has a density of 0.737 g/ml at 25 ∘c. how many grams of this gasoline would fill a 14.3 gal tank (1us gal=3.78l)? express your answer to three significant figures and include the appropriate units.

the tank will be filled with 39,8*10³g of gasoline

Explanation:

Given that the units for density are g/mL, the first step is to know the volume of tank expressed in L.using the value given in the question  

Tank volume (L)= 14.3 gal* (3.78L/1gal)= 54.054 L

Now, we need this volume expressed in mL, Remember 1L = 1000mL, so

Tank volume (mL)= 54.054L * (1000mL/ 1L) = 54054mL

Now, we can use the density to get the answer.

gasoline (g) = 54054mL* 0.737g/mL =  39837g

There are 5 significant figures in 39837, for that reason we need to reduce this expression to get only 3 significant figures, but without altering the result.

If we use scientific notation, we could write 39,837*10³g of gasoline, and finally we rounded this number to get just 3 significant figures, and we get 39.8*10³g of gasoline

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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