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3 years ago

Can someone help me? -just check my latest question.. thanks...

Chemistry

1 answer:

Harman [31]3 years ago

7 0

Explanation:

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Which reagent is the limiting reactant when 0.400 mol Al(OH)3 and 0.400 mol H2SO4 are allowed to react

jeka94

Answer:

H₂SO₄ will be the limiting reagent.

Explanation:

The balanced reaction is:

2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction).

You can use a simple rule of three as follows: if by stoichiometry 2 moles of Al(OH)₃ reacts with 3 moles of H₂SO₄, how much moles of H₂SO₄ will be needed if 0.4 moles of Al(OH)₃ react?

$moles of H_{2} SO_{4} =\frac{0.4 moles of Al(OH)_{3}*3 moles of H_{2} SO_{4} }{2 moles of Al(OH)_{3}}$

moles of H₂SO₄= 0.6 moles

But 0.6 moles of H₂SO₄ are not available, 0.40 moles are available. Since you have less moles than you need to react with 0.4 moles of Al(OH)₃, H₂SO₄ will be the limiting reagent.

7 0

3 years ago

An unknown insoluble substance displaced the water shown. It's mass is indicated on the triple beam balance. Mass = A. 694 B. 69

vodka [1.7K]

Answer:

D. 693.8.

Explanation:

The first scale of the triple beam balance indicate that the mass is 600.

The second scale of the triple beam balance indicate that the mass is 690.

The third scale of the triple beam balance indicate that the mass is higher that 693.5 and the scale is near but lower than 694, so the mass = 693.8.c

3 0

3 years ago

Read 2 more answers

For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be prod

natima [27]

Answer:

(a)

Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of $CS_2$ = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of $H_2S$ = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

(a)

For the first reaction:-

$NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}$

The mole ratio of the reactants = 1 : 1

0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride

So, Moles of ammonium chloride formed = 0.5 moles

Molar mass of ammonium chloride = 53.491 g/mol

Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g

(b)

For the first reaction:-

$CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}$

The mole ratio of the reactants = 1 : 4

It means

0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.

But available moles of S = 0.5 moles

Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

4 moles of S produces 1 mole of $CS_2$

Thus,

0.5 moles of S produces $\frac{1}{4}\times 0.5$ mole of $CS_2$

Mole of $CS_2$ = 0.125 mole

Molar mass of $CS_2$ = 76.139 g/mol

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

4 moles of S produces 2 moles of $H_2S$

Thus,

0.5 moles of S produces $\frac{2}{4}\times 0.5$ mole of $H_2S$

Mole of $H_2S$ = 0.25 mole

Molar mass of $H_2S$ = 34.1 g/mol

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

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3 years ago

Electron Identify the type of charge of the following particles in an

Alexeev081 [22]

Answer:

Electron charge: -1

Neutron charge: 0

Proton charge: +1

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3 years ago

Is a Animal and Plant cell living

Mila [183]

Yes Animal and plant are cell living

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3 years ago

Can someone help me? -just check my latest question.. thanks... ​

Can someone help me? -just check my latest question.. thanks...