Question

Please I really really need help!!!

Answer 1

Answer:

1) 40

2) 2.25 moles

3) 17

4) 120

5) Fe₂O₃

Explanation:

Please see attached picture for full solution.

Answer 2

Answer:

1) $\boxed{Molar \ mass = 40\ g/mol}$

2) $\boxed{No.\ of\ Moles = 1.61 \ moles}$

3) $\boxed{Ammonia = 16\ g/mol}$

4) $\boxed{Sodium\ Hydrogen\ Sulphate= 120 \ g/mol}$

5) $\boxed{Fe_{2}O_{3}}$

Explanation:

Question # 1:

Using Formula, No. of moles = Mass in grams / molar mass

Where moles = 0.375 mol. , mass = 15 g

=> Molar Mass = Mass in grams / No. of moles

=> Molar Mass = 15 / 0.375

=> Molar mass = 40 g/mol

Question # 2:

No. of moles = Mass in grams / Molar mass

Where Mass = 90 g ,

=> Molar Mass = $MgO_{2}$

=> MM = 24 + (16*2)

=> MM = 24 + 32

=> MM = 56 g/mol

No. of Moles = 90 / 56

No. of Moles = 1.61 moles

Question # 3:

Ammonia => $NH_{3}$

N has atomic mass 14 ang H has atomic mass 1

Ammonia = (14)+(1*2)

Ammonia = 14+2

Ammonia = 16 g/mol

Question # 4:

Sodium Hydrogen Sulphate => $NaHSO_{4}$

Where Na has atomic mass 23, H has atomic mass 1 , sulpher 32 and oxygen 16

$NaHSO_{4}$ = 23+1+32+(16*4)

=> 56+64

=> 120 g/mol

Question # 5:

The metal which has relative atomic mass of 56 is Iron (Fe)

Given that the oxide contains 70.0 % of Metal

Mass of $MO_{x}$ (Metal oxide) = 56 / 70 * 100

Mass of $MO_{x}$ (Metal oxide)  = 0.8 * 100

Mass of $MO_{x}$ (Metal oxide)  = 80

Mass of x O's = 80 - 56

Mass of x O's = 24

Now, Mass of O = 24 / 16

Mass of O = 1.5

So, Fe Metal and its oxide are in the ratio of

1 : 1.5

×2   ×2

2 : 3

So, the empirical formula of metal with its oxide is $Fe_{2}O_{3}$