Question

two mteallic oxide conation 27.6% and 30% oxygen respectively if the formula of the first oxide is x3o4 that of the second will be

Answer 1

Answer:

The formula of the second oxide is :$x_1O_{1.5}=x_2O_3$

Explanation:

Percentage of oxygen in in first oxide = 27.6%

Then the percentage of element x will be = 100% - 27.6 % = 73.3 %

Formula of first oxide:$X_3O_4$

Mass of the $x_3O_4$ be M

$27.6\%=\frac{4\times 16 g/mol}{M}\times 100$

M = 231.88 g/mol

Mass of element x present in $X_3O_4=M_x$

$73.3\%=\frac{M_x}{231.88 g/mol}\times 100$

$M_x=169.96 g/mol$

Mass of single element x =$\frac{169.96 g/mol}{3}=56.65 g/mol$

Percentage of  oxygen in second oxide = 30%

Then the percentage of element x will be = 100% - 30% = 70 %

Moles of oxygen atom in 30 g =$\frac{30 g}{16 g/mol}=1.875 mol$

Moles of x atom in 70 g = $\frac{70 g}{56.65 g/mol}=1.235 mol$

Dividing number moles of each from the smallest value of moles.

For Oxygen = $\frac{1.875 mol}{1.235 mol}=1.51\approx 1.5$

For element x = $\frac{1.235 g}{1.235 g}=1$

The formula of the second oxide is :$x_1O_{1.5}=x_2O_3$