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Vinil7 [7]
2 years ago
11

2. If a student drops a 2.3 g piece of magnesium into a flask of hydrochloric acid, this reaction occurs: Mg + 2HCl MgCl2 + H2

Chemistry
1 answer:
mezya [45]2 years ago
6 0

1.2 L of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Step 1: Write the balanced equation

Mg + 2 HCl ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 2.3 g of Mg

The molar mass of Mg is 24.31 g/mol.

2.3 g × 1 mol ÷24.31 g = 0.095 mol

Step 3: Calculate the moles of H₂ produced

0.095 mol Mg × 1 mol H₂ ÷ 1 mol Mg = 0.095 mol H₂

Step 4: Calculate the volume occupied by the hydrogen

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T÷P

V = 0.095 mol × (0.0821 atm.L/mol.K) × 298 K÷2 atm

V =  1.2 L

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

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(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe
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Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration C_o = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

time (t) = \dfrac{V}{Q}

time (t) = \dfrac{200}{0.3}

time (t) = 666.66 \ sec

time (t) = \dfrac{666.66 }{3600} hrs

time (t) = 0.185 hrs

\text{Using First Order Reaction:}

\dfrac{dc}{dt}=kc

where;

t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)

0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)

0.942 =  \Big ( \dfrac{200}{C_e}- 1 \Big)

1+ 0.942 =  \Big ( \dfrac{200}{C_e} \Big)

\dfrac{200}{C_e} = 1.942

C_e = \dfrac{200}{1.942}

\mathbf{C_e = 102.98 \ mg/l}

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)

\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)

\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)

V_{PFR} =0.196(0.663)*0.3*3600

V_{PFR} = 140.34 \ m^3

The volume of the PFR is ≅ 140 m³

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