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Annette [7]
3 years ago
11

What is the mass of solid NH4Cl formed when 75.5g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas remai

ning, measured at 14.0C and 752 mmHg? What gas is it? The formula is NH3 (g) + HCl gas -> NH4Cl solid
Chemistry
2 answers:
taurus [48]3 years ago
8 0

Answer:The volume of the remaining gas that is ammonia is 23.85 L.

Explanation:

NH_3(g)+HCl(g)\rightarrow NH_4Cl(s)

Moles of NH_3=\frac{\text{mass of} NH_3}{\text{Molar mass of}NH_3}=\frac{75.5 g}{17.03 g/mol}=4.43 mole

Moles of HCl of gas = \frac{\text{mass of} HCl}{\text{Molar mass of}HCl}=\frac{75.5 g}{36.5 g/mol}=2.06 mol

According to reaction 1 mole of HCl reacts with 1 mol of NH_3 then 2.06 moles of HCl will react with = 2.06 moles of NH_3

Moles left of ammonia left = 4.43 - 2.06 = 2.36 moles

Volume of the gas will be given by Ideal gas equation: PV=nRT

Pressure = 752 mmHg = 752 × 0.0031 atm = 2.33 atm

R = 0.08026 L atm/K mol

V = ? , n = number of moles of ammonia

Temperature = 14 °C = 14 + 273 K = 287 K(0°C = 273K)

V=\frac{2.36 mol\times 0.08206 L atm/K mol\times 287 K}{2.33 atm}=23.85 L

The volume of the remaining gas that is ammonia is 23.85 L.

weqwewe [10]3 years ago
5 0

Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

Explanation :

The balanced chemical reaction will be:

NH_3+HCl\rightarrow NH_4Cl

First we have to calculate the moles of NH_3 and HCl.

\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}

Molar mass of NH_3 = 17 g/mole

\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole

and,

\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}

Molar mass of HCl = 36.5 g/mole

\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NH_3

So, 2.07 mole of HCl react with 2.07 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV=nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm     (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 14.0^oC=273+14.0=287K

Putting values in above equation, we get:

0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K

V=56.5L

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

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