Question

How many grams of NH3 are needed to provide the same number of molecules as in 0.85 grams of SF6?

Answer 1

Answer:

0.099 grams of ammonia  are needed to provide the same number of molecules as in 0.85 grams of sulfur hexafluoride.

Explanation:

Mass of sulfur hexafluoride = 0.86 g

Moles of sulfur hexafluoride =$\frac{ 0.86 g}{146 g/mol}=0.005822 mol$

1 mole = $N_A=6.022\times 10^{23} molecules/ atom$

This means 0.005822 moles of ammonia will have same number of molecules as that of the molecules in 0.005822 moles of sulfur hexafluoride .

Moles of ammonia = 0.005822 mole

Mass of 0.005822 moles of ammonia = $0.005822 mol\times 17 g/mol=0.099 g$

0.099 grams of ammonia  are needed to provide the same number of molecules as in 0.85 grams of sulfur hexafluoride.

Answer 2

Above it says the molecular weights are

NH3- 17g/mol and SF6-146 g/mol

Well 1 mole of SF6 is 146.048 grams (i added hte atomic masses of each element). So then the number of moles in 0.85 grams would be 0.00582000438 moles.

= 1mole / 146.048g * 0.85g

so we would need 0.00582000438 moles of NH3 to have the same number of molecules.

One mole of NH3 is 17.030519999989988 grams (i added each atoms mass). so 0.00582000438 moles of NH3 would be:

= 17.030519999989988 g / mole * 0.00582000438moles

that equals 0.09911770099 grams.

so 0.09911770099 grams is the answer if you round that you get about 0.1 grams