How many grams of f are in 12.56 g of sf6? h.

9.80 g

Explanation:

The molecular mass of the atoms mentioned in the question is as follows -

S = 32 g / mol

F = 19 g / mol

The molecular mass of the compound , SF₆ = 32 + ( 6 * 19 ) = 146 g / mol

The mass of 6 F = 6 * 19 = 114 g /mol .

The percentage of F in the compound =

mass of 6 F / total mass of the compound * 100

Hence ,  

The percentage of F in the compound = 114 g /mol  / 146 g / mol * 100

78.08 %

Hence , from the question ,

In 12.56 g of the compound ,

The grams of F = 0.7808 * 12.56 = 9.80 g

400 cm³ of ammonia, NH₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.

Finally, we shall determine the maximum volume of ammonia, NH₃ produced from the reaction. This can be obtained as illustrated below:

From the balanced equation above,

3 cm³ of H₂ reacted to produce 2 cm³ of NH₃.

Therefore, 600 cm³ of H₂ will react to produce = (600 × 2)/3 = 400 cm³ of NH₃.

Thus, 400 cm³ of ammonia, NH₃ were obtained from the reaction.