Respuesta :
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 [tex]YCl_3[/tex].
1.835 grams of [tex]YCl_3[/tex]can be prepared.
Explanation:
[tex]2M+3X_2\rightarrow 2MX_3[/tex]
Moles of [tex]X_2[/tex] =n
Number of moleules of [tex]X_2=8.92\times 10^{20} molecules[/tex]
1 mole = [tex]6.022\times 10^{23} molecules[/tex]
[tex]n=\frac{8.92\times 10^{20} molecules}{6.022\times 10^{23} molecules}[/tex]
n = 0.001481 mole
Mass of [tex]X_2=0.105 g[/tex]
Molar mass of [tex]X_2=m[/tex]
[tex]n=\frac{Mass}{\text{Molar mass}}[/tex]
[tex]0.001481 mol=\frac{0.105 g}{m}[/tex]
m = 71 g/mol
Atomic mass of X = [tex]\frac{71 g/mol}{2}=35.5 g/mol[/tex]
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 :
[tex]=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100[/tex]
X:
[tex]54.47\%=\frac{3\times 35.5 g/mol}{M'}\times 100[/tex]
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 [tex]YCl_3[/tex].
[tex]2Y+3Cl_2\rightarrow 2YCl_3[/tex]
Moles of Yttrium = [tex]\frac{1g }{89.02 g/mol}=0.01123 mol[/tex]
Moles of chlorine gas= [tex]\frac{1 g}{71 g/mol}=0.01408 mol[/tex]
According to reaction, 3 moles of chlorine reacts with 2 moles of Y.
Then 0.01408 moles of chlorine gas will :
[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] 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 [tex]YCl_3[/tex]
Then 0.01408 moles of chlorine will give :
[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] of [tex]YCl_3[/tex]
Mass of 0.009387 moles of [tex]YCl_3[/tex]:
0.009387 mol × 195.52 g/mol = 1.835 g
1.835 grams of [tex]YCl_3[/tex]can be prepared.
