Question

How many Na+ ions are contained in 5.43 g of Na3PO4?

Answer 1

Answer:

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Explanation:

Explanation:

Sodium sulfate is

N

a

2

S

O

4

First, calculate the moles of sodium sulfate you have in a 14.2g sample.

Find the molar mass of sodium sulfate

molar mass = atomic mass in grams

2(23) + 32 + 4(16) = 142 grams per mole

Divide the mass of the sample by the molar mass to obtain moles

14.2

g

142

g

m

o

l

−

1

= 0.1 mol

Now, for every 1 mol of

N

a

2

S

O

4

you have 2 moles of

N

a

+

ions. Multiple the moles of

N

a

2

S

O

4

that you found just then by 2 to obtain the moles of sodium ions.

Use Avogadro's number (6.022 x

10

23

) to find the number of ions present. (Recall that Avogrado's number is the number of particles per mole of a substance).

6.022 x

10

23

x 0.2 mol = 1.2044 x

10

23

ions

Answer 2

The Na+ ions contained in 5.43 g of Na₃PO₄ are 0.099.

Given:

5.43 g of Na₃PO₄.

To Find:

The Na+ ions are contained in 5.43 g of Na₃PO₄.

Solution:

To find the Na+ ions present in 5.43g of sodium phosphate (Na₃PO₄) we will follow the following steps:

As we know,

Moles =

$\frac{given \: mass}{molecular \: mass}$

According to the question:

Given mass = 5.43g

The molecular mass of sodium phosphate = sum of the mass of (3 sodium + 1 phosphorous + 4 oxygen)

So,

The molecular mass of sodium phosphate =

$3 \times 23 +31 +4 \times 16 = 69 +31 +64 = 164g$

So,

Moles =

$\frac{5.43}{164} = 0.033$

Now,

1 mole of sodium phosphate gives 3 sodium ions.

So,

0.033 moles will give 0.033 × 3 sodium ions.

Total sodium ions in 5.43 g of sodium phosphate = 0.099

Henceforth, The Na+ ions contained in 5.43 g of Na₃PO₄ is 0.099.

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