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3 years ago

Use the following reaction:

Chemistry

2 answers:

Amiraneli [1.4K]3 years ago

7 0

C. 1.96 grams 02 hope this helpsss ;)

KengaRu [80]3 years ago

3 0

Sport but I need some points

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Calculate thr number of photons having a wavelength of 10.0 μm required to produce 1.0 kJ of energy.

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2 years ago

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3 years ago

A 25 mL sample of 0.100 M HNO3 completely reacts with NaOH according to this equation:

Anastasy [175]

It is a good thing that you already have answered the first question. Now, moving on to the second question, there exist an equation for the neutralization of acid by a base that is shown below,
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3 years ago

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Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur

yuradex [85]

Answer: The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

$\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol$

For iron(III) chloride:

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol$

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

$3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe$

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = $\frac{2}{3}\times 1.708=1.114mol$ of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

$0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g$

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

6 0

3 years ago