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3 years ago

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Given the balanced chemical equation, SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH°rxn = −184 kJ Determine the mass (in grams) of H

F(g) must react when 345 kJ of energy is released.

1 answer:

Ostrovityanka [42]3 years ago

3 0

Answer:

mass HF = 150.05 g

Explanation:

SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

⇒ Q = (ΔH°rxn * mHF) / (mol HF * MwHF )

∴ MwHF = 20.0063 g/mol

∴ mol HF = 4 mol

∴ ΔH°rxn = - 184 KJ

∴ Q = 345 KJ

mass HF ( mHF ):

⇒ mHF = ( Q * mol HF * MwHF ) / ΔH°rxn

⇒ mHF = ( 345 KJ * 4mol HF * 20.0063 g/mol ) / 184 KJ

⇒ mHF = 150.05 g HF

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Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

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3 years ago

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from

IgorLugansk [536]

The question is incomplete, here is the complete question.

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g $S_8$ .

Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.

Explanation : Given,

Mass of Ag = 3.0 g

Mass of $S_8$ = 3.0 g

Molar mass of Ag = 107.8 g/mole

Molar mass of $S_8$ = 256 g/mole

Molar mass of $Ag_2S$ = 247.8 g/mole

First we have to calculate the moles of Ag and $S_8$ .

$\text{ Moles of }Ag=\frac{\text{ Mass of }Ag}{\text{ Molar mass of }Ag}=\frac{3.0}{107.8g/mole}=0.0278moles$

$\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)$

From the balanced reaction we conclude that

As, 16 mole of $Ag$ react with 1 mole of $S_8$

So, 0.0278 moles of $Ag$ react with $\frac{0.0278}{16}=0.00174$ moles of $S_8$

From this we conclude that, $S_8$ is an excess reagent because the given moles are greater than the required moles and $Ag$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag_2S$

From the reaction, we conclude that

As, 16 mole of $Ag$ react to give 8 mole of $Ag_2S$

So, 0.0278 moles of $Ag$ react to give $\frac{0.0278}{16}\times 8=0.0139$ moles of $Ag_2S$

Now we have to calculate the mass of $Ag_2S$

$\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S$

$\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g$

Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.

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