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3 years ago

What are the coefficients that balance the following chemical equation? _Ag + _ S → ___Ag2S

Chemistry

2 answers:

Tanya [424]3 years ago

4 0

Answer:

$2Ag+S$ → $Ag_2S$

Explanation:

In order to balance this, we want to have the same number of elements on each side of the equation.

On the right side, we see that we have two atoms of Ag and one atom of S. So on the left, we want to have 2 atoms of Ag and 1 atom of S. We can put 1 in front of S and 2 in front of Ag:

2Ag + 1S → 1 $Ag_2S$

We actually don't have to explicitly write the "1":

$2Ag+S$ → $Ag_2S$

Hope this helps!

Andrej [43]3 years ago

3 0

Answer:

1Ag+2S⇔1Ag2S

Explanation:

The first step is to see how many of each element are on each side of the equation. There is one silver on the left and one silver on the right, so you can leave those alone. However, there is one sulfur on the left and two on the right. You cannot multiply the right by 2 because then that would upset the balance of the silver, but you can multiply the sulfur by 2. Therefore, the equation should be 1Ag+2S⇔1Ag2S. Hope this helps!

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Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

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Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

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The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

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3 years ago

What are the coefficients that balance the following chemical equation? ___Ag + ___ S → ___Ag2S

What are the coefficients that balance the following chemical equation? _Ag + _ S → ___Ag2S