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Helga [31]
2 years ago
6

If 0.600mol of chloride gas reacted with 0.500mol of aluminium metal to produce aluminium chloride,which reactant is in excess?h

ow many moles of aluminium chloride can be produced during the reaction.
Chemistry
1 answer:
Rus_ich [418]2 years ago
4 0

Answer:

The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3

Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3

Given: 0.600 moleCl2; 0.500 mole Al

Required: Excess reactant___; Number of moles of AlCl3 produced__

Solution: Use dimensional analysis using the mole-mole rel

0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al

0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2

Based on the given:

0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)

0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)

Answer: Excess reactant is Al; Limiting reactant is Cl2

The amount of AlCl3 produced = 0.6 mole Cl2 + 0.4 mole Al = 1.0 mole AlCl3

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Zinaida [17]
A solid formed from liquid reactants chemical reactions is called a precipitate.
3 0
3 years ago
Read 2 more answers
Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.
Rudiy27
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

                                ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where: 
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3

Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol

m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg


For the problem, 
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius
5 0
3 years ago
Answer all questions pls
tekilochka [14]

Answer:

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Explanation:

U dont have to copy and paste this put these are some ideas to use for ur answers

6 0
2 years ago
2. Based on what you know about waves and light, do you think that light can be
ss7ja [257]

Answer:

Light as a wave: Light can be described (modeled) as an electromagnetic wave. In this model, a changing electric field creates a changing magnetic field. This changing magnetic field then creates a changing electric field and BOOM - you have light. ... So, Maxwell's equations do say that light is a wave.

Explanation:

Hope this helps

6 0
2 years ago
Read 2 more answers
In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
2 years ago
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