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8090 [49]
2 years ago
14

Boron trifluoride gas is collected at 2.0 degree C in an evacuated flask with a measured volume of 15.0 L. When all the gas has

been collected, the pressure in the flask is measured to be 0.130 atm. Calculate the mass and number of moles of boron trifluoride gas that were collected. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
IgorC [24]2 years ago
7 0

Answer:

Moles of boron trifluoride gas that were collected = 11.6 mol

Mass of boron trifluoride gas that were collected = 787 g

Explanation:

Given that:

Temperature = 2.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (2.0 + 273.15) K = 275.15 K

V = 15.0 L

Pressure = 0.130 atm

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

0.130 atm × 15.0 L = n ×0.0821 L atm/ K mol  × 275.15 K

⇒n = 11.6 mol

<u>Thus, Moles of boron trifluoride gas that were collected = 11.6 mol </u>

Molar mass of boron trifluoride gas = 67.82 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

11.6\ mole= \frac{Mass}{67.82\ g/mol}

<u>Mass of boron trifluoride gas that were collected = 787 g</u>

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Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of c
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The question is incomplete. The complete question is

Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed

Answer:

a)84.91g

b)8.20g

c)316.4g

d)616.73g

Explanation:

The equation of the reaction:

2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)

Molar mass of potassium phosphate= 212.27 g/mol

Amount of potassium phosphate= 500/212.27= 2.4 moles

Molar mass of calcium nitrate= 164.088 g/mol

Amount of calcium nitrate= 500/164.088=3.05moles

a) amount of potassium phosphate reacted according to reaction equation= 2 moles

Amount of potassium phosphate remaining= 2.4-2=0.4 moles

Mass of potassium phosphate remaining= 0.4×212.27=84.91g

b) Amount of calcium nitrate reacted according to reaction equation=3

Amount of calcium nitrate remaining=3.05-3= 0.05

Mass of calcium nitrate remaining= 0.05×164.088= 8.20g

c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.

From the reaction equation,

3 moles of calcium nitrate yields 1 mole of calcium phosphate

3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate

Molar Mass of calcium phosphate= 310.18 g/mol

Mass of calcium phosphate produced= 1.02×310.18= 316.4g

d)

3 moles of calcium nitrate yields 6 moles of potassium nitrate

3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate

Molar mass of potassium nitrate = 101.1032 g/mol

Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g

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