Question

An ionic compound MX3 is prepared according to the following unbalanced chemical equation. M + X2 gives MX3, A 0.105-g sample of X2 contains 8.92 X 10^20 molecules. The compound MX3 consists of 54.47% X by mass. What are the identities of M and X, and what is the correct name for MX3? Starting with 1.00 g each of M and X2, what mass of MX3 can be prepared?

Answer 1

Atomic mass of 35.5 g/mol is of chlorine.

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for
`YCl_3`.

1.835 grams of
`YCl_3`can be prepared.

Step-by-step explanation:

`2M+3X_2\rightarrow 2MX_3`

Moles of
`X_2` =n

Number of moleules of
`X_2=8.92* 10^(20) molecules`

1 mole =
`6.022* 10^(23) molecules`

`n=(8.92* 10^(20) molecules)/(6.022* 10^(23) molecules)`

n = 0.001481 mole

Mass of
`X_2=0.105 g`

Molar mass of
`X_2=m`

`n=\frac{Mass}{\text{Molar mass}}`

`0.001481 mol=(0.105 g)/(m)`

m = 71 g/mol

Atomic mass of X =
`(71 g/mol)/(2)=35.5 g/mol`

Atomic mass of 35.5 g/mol is of chlorine.

The compound MX3 consists of 54.47% X by mass:

Molar mass of compound = M'

Percentage of element in compound :

`=\frac{\text{number of atoms}* text{Atomic mass}}{\text{molar mas of compound}}* 100`

X:

`54.47\%=(3* 35.5 g/mol)/(M')* 100`

M' = 195.52 g/mol

Molar mass of compound = M'

M' = 1 × (atomic mass of M)+ 3 × (atomic mass of X)

195.52 g/mol = atomic mass of M + 3 × (35.5 g/mol)

Atomic mass of M = 89.02 g/mol

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for
`YCl_3`.

`2Y+3Cl_2\rightarrow 2YCl_3`

Moles of Yttrium =
`(1g )/(89.02 g/mol)=0.01123 mol`

Moles of chlorine gas=
`(1 g)/(71 g/mol)=0.01408 mol`

According to reaction, 3 moles of chlorine reacts with 2 moles of Y.

Then 0.01408 moles of chlorine gas will :

`(2)/(3)* 0.01408 mol=0.009387 mol` of Y.

This means that chlorine is in limiting amount., So, amount of yttrium (III) chloride will depend upon amount of chlorine gas.

According to reaction , 3 moles of chlorine gives 2 moles of
`YCl_3`

Then 0.01408 moles of chlorine will give :

`(2)/(3)* 0.01408 mol=0.009387 mol` of
`YCl_3`

Mass of 0.009387 moles of
`YCl_3`:

0.009387 mol × 195.52 g/mol = 1.835 g

1.835 grams of
`YCl_3`can be prepared.