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3 years ago

How many grams of ferric oxide will you need to make 1.5 M solution?

Chemistry

1 answer:

Masteriza [31]3 years ago

3 0

Answer:Stoichiometry calculations are about calculating the amounts of substances that react and form in a

chemical reaction. For example, based on the balanced chemical equation, we can calculate the

amount of a product substance that will form if we begin with a specific amount of one or more

reactants. Or, you may have a target amount of product to prepare. How much starting compounds

are needed to prepare this amount? These are practical calculations that are done frequently by

chemists.

The Method

For practically all stoichiometry problems, we want to use the. . . .

Fabulous Four Steps

Step 1: Write the balanced chemical equation for the reaction.

Step 2: Calculate the moles of "given" substance. If more than one reactant amount is given,

calculate the moles of each to determine which is the limiting reactant.

Step 3: Calculate the moles of "desired" substance from your answer in Step 2 using the

coefficients from the balanced chemical equation. If more than one reactant was given originally, you

can calculate the moles of product twice, based on the moles of each reactant. The reactant that

gives the smaller moles of product is the limiting reactant. Keep this answer for Step 4.

Step 4: Convert your answer in Step 3 to the units the problem asks for. Usually this is grams,

but it could be volume (for gases or liquid solutions) or concentration (such as molarity, for solutions).

Again in brief:

1. Balanced reaction

2. Moles of "given" substance(s)

3. Moles of "desired" substance such as a product

4. Convert Step 3 answer to the units asked fo

Explanation:

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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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