How many moles are there in 397 grams of #Na_2SO_4#?

Answer 1

2.80

A mole, which is determined for each molecule by adding the individual atomic weights of its constituent atoms, is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms.

For #Na_2SO_4# this is #2 x 23 = 46# for the sodium, plus #4 x 16 = 64# for the oxygen, plus #32# for the sulfur. That adds up to a molecular weight of #142 (g/"mole")#. Now we divide the given number of grams by this value to derive the number of moles. #(397/142)(g/(g/"mol")) = 2.80 "moles"#

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Answer 2

Wyatt Atkins

#n_(Na_2SO_4)=2.79 mol#

Molar mass of #Na_2SO_4 =142.04g/(mol)#

Apply the formula,

#n_(Na_2SO_4)=m/M#

wherein

#n=# number of moles #m=#mass (in grams) #M=#molar mass (in #(grams)/(mol)#)

#n_(Na_2SO_4)=(397g)/(142.04(g/(mol)))#

#n_(Na_2SO_4)=2.79 mol#

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Answer 3

Avery Adams

To find the number of moles in 397 grams of Na2SO4, divide the given mass by the molar mass of Na2SO4, which is approximately 142 grams per mole.

( \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{397 , \text{g}}{142 , \text{g/mol}} \approx 2.79 , \text{moles})

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.