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

Answer:

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.

Explanation:

$2M+3X_2\rightarrow 2MX_3$

Moles of $X_2$ =n

Number of moleules of $X_2=8.92\times 10^{20} molecules$

1 mole = $6.022\times 10^{23} molecules$

$n=\frac{8.92\times 10^{20} molecules}{6.022\times 10^{23} molecules}$

n = 0.001481 mole

Mass of $X_2=0.105 g$

Molar mass of $X_2=m$

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

$0.001481 mol=\frac{0.105 g}{m}$

m = 71 g/mol

Atomic mass of X = $\frac{71 g/mol}{2}=35.5 g/mol$

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 :

$=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100$

X:

$54.47\%=\frac{3\times 35.5 g/mol}{M'}\times 100$

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 = $\frac{1g }{89.02 g/mol}=0.01123 mol$

Moles of chlorine gas= $\frac{1 g}{71 g/mol}=0.01408 mol$

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

Then 0.01408 moles of chlorine gas will :

$\frac{2}{3}\times 0.01408 mol=0.009387 mol$ 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 :

$\frac{2}{3}\times 0.01408 mol=0.009387 mol$ 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.