Question

${\mathit{M}}_{\mathbf{2}}{\mathit{S}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$When is heated in air, it is converted to.$\mathit{M}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$A 4.000 $\mathit{g}$sample of ${\mathit{M}}_{\mathbf{2}}{\mathit{S}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$shows a decreases in mass of $\mathbf{0}\mathbf{.277}\mathit{g}$when it is heated in air. What is the average atomic massof.$M$

Short answer

The average atomic mass for$M$ is.$184u$

Step-by-step solution

M2S3Reaction towards air

$M_2{S}_3+5O_2\to 2M{O}_2+3S{O}_2$

When$M_2{S}_3$ is reacted with ir it forms$M{O}_2$ its mass is decreases to $0.277g$. Therefore the mass of $2M{O}_2$is formed to$3.723g$

Mass value of reaction

Mass of$2M{O}_2$$=3.723g$

Mass of$M_2{S}_3=4g$$x$

To find the Mass value assume that$M$ is

Calculate the moles

Moles of$M_2{S}_3$$=\frac{4g}{(2x+96.2)g/\text{mol}}$

Moles of$2M{O}_2$$=\frac{3.723g}{(2x+64)g/\text{mol}}$

Based on the reaction the$M_2{S}_3$ is converted to$M{O}_2$

Conclusion

As per the reaction

$\text{Moles\hspace{0.17em}of\hspace{0.17em}}{M}_2{S}_3$$=\text{\hspace{0.17em}Moles\hspace{0.17em}of\hspace{0.17em}}2M{O}_2$

$\frac{4g}{(2x+96.2)g/\text{mol}}=\frac{3.723g}{(2x+64)g/\text{mol}}$

The average atomic mass for$M$is$184u$$184u$