Question

1.16 gram of metal M reacts with excess of O2 and convert all Metal M intoM3O2 the maximum amount of M3O2vproduced is atomic mass of M is equal to 58 g/mol​

Answer 1

Answer:

The maximum amount of $M_3O_2$ produced is 1.3596 g.

Explanation:

Atomic mass of metal m = 58 g/mol

Moles of metal M = $\frac{1.16 g}{58 g/mol}=0.02 mol$

$3M+O_2\rightarrow M_3O_2$

According to reaction , 3 moles of metal M gives 1 moles of $M_3O_2$

Then 0.02 moles of metal  M will give:

$\frac{1}{3}\times 0.02=0.0066$ moles of $M_3O_2$

Molecular mass of of $M_3O_2=3\times 58 g/mol+2\times 16=206 g/mol$

Mass of 0.0066 moles of $M_3O_2=0.0066 mol\times 206 g/mol=1.3596 g$

The maximum amount of $M_3O_2$ produced is 1.3596 g.

Answer 2

Explanation:

The maximum amount of M_3O_2M

3

O

2

produced is 1.3596 g.

Explanation:

Atomic mass of metal m = 58 g/mol

Moles of metal M = \frac{1.16 g}{58 g/mol}=0.02 mol

58g/mol

1.16g

=0.02mol

3M+O_2\rightarrow M_3O_23M+O

2

→M

3

O

2

According to reaction , 3 moles of metal M gives 1 moles of M_3O_2M

3

O

2

Then 0.02 moles of metal M will give:

\frac{1}{3}\times 0.02=0.0066

3

1

×0.02=0.0066 moles of M_3O_2M

3

O

2

Molecular mass of of M_3O_2=3\times 58 g/mol+2\times 16=206 g/molM

3

O

2

=3×58g/mol+2×16=206g/mol

Mass of 0.0066 moles of M_3O_2=0.0066 mol\times 206 g/mol=1.3596 gM

3

O

2

=0.0066mol×206g/mol=1.3596g

The maximum amount of M_3O_2M

3

O

2

produced is 1.3596 g.