Answer:
96.33%
Explanation:
The equation of the reaction is represented as:
[tex]Ni_{(s)} + 4CO_{(g)}[/tex] ⇒ [tex]Ni(CO_{4})_{(g)}[/tex]
the molar mass of Ni = 58.7 g/mol
the molar mass of [tex]Ni(CO_{4})_{(g)}[/tex] = 171 g/mol
At Standard Temperature and Pressure;
25.0 L of CO = [tex]\frac{25.0}{22.4}[/tex] =1.116mol
6.25 L of [tex]Ni(CO_{4})_{(g)}[/tex] = [tex]\frac{6.25}{22.41}[/tex] = 0.279mol
To calculate for the percentage of Ni in the metal sample, we need to keep CO in excess, thereafter, we can say 0.279mol mol of Ni(s) equals to 0.279 mol of [tex]Ni(CO_{4})_{(g)}[/tex].
i.e
[tex]Ni(CO_{4})_{(g)}[/tex], in excess will require = [tex]\frac{1}{4}*1.116[/tex] = 0.279 mol of Ni(s).
Given that, molar mass of Ni = 58.7 g/mol
and no of moles = [tex]\frac{mass}{molar mass}[/tex]
the mass can be calculated as = no of moles × molar mass
= 0.279 × 58.7
= 16.377 g Ni present in the sample
∴ the percentage of Ni in the metal sample = [tex]\frac{16.377}{17.0}*100%[/tex]
=96.33%
The percentage of Ni in the metal sample has been 96.33%.
The balanced equation for the reaction:
[tex]\rm Ni\;+\;4\;CO\;\rightarrow\;Ni(CO_4)[/tex]
At STP:
22.4 L gas = 1 mole
25 L carbon dioxide = 1.116 mol
6.25 L [tex]\rm Ni(CO_4)[/tex] = 0.279 mol.
1 mole [tex]\rm Ni(CO_4)[/tex] = 1 mole Ni
0.279 mol [tex]\rm Ni(CO_4)[/tex] =0.279 mol Ni.
Mass = moles × molecular weight
Mass of Ni = 0.279 [tex]\times[/tex] 58.7 g/mol
Mass of Ni = 16.377 g
The sample nickel = 17 grams.
Percentage of Nickel = [tex]\rm \dfrac{16.377}{17}\;\times\;100[/tex]
Percentage of Nickel = 96.33%.
The percentage of Ni in the metal sample has been 96.33%.
For more information about the percentage sample, refer to the link:
https://brainly.com/question/19637992
