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2 years ago

How many grams of NH3 are required to produce 4.65 g of HF

Mathematics

1 answer:

nignag [31]2 years ago

8 0

1.326 grams of NH3 are required to produce 4.65 g of HF.

Step-by-step explanation:

Balanced chemical reaction is written first to know the number of moles taking part in original reaction.

NH3 +3 F2 ⇒3 HF + NF3

Given:

mass of HF = 4.65

First the number of moles of HF in 4.65 grams is calculated by using the formula:

number of moles (n) = $\frac{mass}{atomic mass of 1 mole}$

atomic mass of HF = 20 grams/mole

putting the values in the above equation number of moles can be found.

n = $\frac{4.65}{20}$

= 0.235 moles of HF are given.

From the equation it can be said that:

1 mole of NH3 reacts to form 3 moles of HF

so, x moles of NH3 would react to form 0.235 moles of HF

$\frac{3}{1}$ = $\frac{0.235}{x}$

3x = 0.235

x = $\frac{0.235}{3}$

x = 0.078 moles of NH3 is required.

The moles are converted to mass by applying the formula:

mass = atomic mass X number of moles (atomic mass of NH3 = 17 grams/mole)

putting the values in the formula

mass = 17 X 0.078

mass = 1.326 grams

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