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[05.04] How many moles of NaF can be produced when 15.0 L F2 gas at 295 K and 1.10 atm react with excess NaCl?F2(g) + 2NaCl(s) →→→→Cl₂(g) + 2NaF (s)

Respuesta :

In this question, we have the following reaction:

F2 + 2 NaCl --> Cl2 + 2 NaF

We have:

15.0 L of F2

295 K

1.10 atm of pressure

In order to find the number of moles of NaF, we need to find the number of moles of F2, we will be using the Ideal gas law to do that:

PV = nRT

R is the gas constant 0.082

Adding the values into the formula:

1.10 * 15 = n * 0.082 * 295

16.5 = 24.19n

n = 16.5/24.19

n = 0.68 moles of F2

Since we have 0.68 moles of F2 and the molar ratio between F2 and NaF is 1:2, 1 mol of F2 to every 2 moles of NaF, therefore:

1 F2 = 2 NaF

0.68 F2 = x NaF

x = 2 * 0.68

x = 1.36 moles of NaF

We will have 1.36 moles of NaF

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