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Alchen [17]
3 years ago
11

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

Mathematics
1 answer:
nignag [31]3 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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