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aksik [14]
3 years ago
7

What volume of H2S is needed to produce 14.2 L of SO2 gas? The balanced equation is: 2H2S + 3O2 → 2SO2 + 2H2O.

Chemistry
1 answer:
storchak [24]3 years ago
5 0

Answer:

14.2 L H2S

Explanation:

Mole ratio

2 mol H2S:2 mol SO2

We are just looking for volume, so the required amount is the same as the produced amount as long as the mole ratio is balanced.

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Xelga [282]
The circulatory system
8 0
3 years ago
What is the molar ratio for the following equation after it has been properly balanced?
hichkok12 [17]

6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe

<h3>Explanation</h3>

Method One: Refer to electron transfers.

Oxidation states:

  • Na: from 0 to +1; loses one electron.
  • Fe: from +3 to 0; gains three electrons.

Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.

6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe

  • There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
  • There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.

6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

Method Two: Atoms conserve.

Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume <em>one</em> as its coefficient.

? Na + <em>1</em> Fe₂O₃ → ? Na₂O + ? Fe

There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.

? Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe

Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.

<em>6</em> Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe

Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.

6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

6 0
3 years ago
Determine the percent water in MgSO4*7H20?
Semmy [17]

Answer:

\% H_2O=51.2\%

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol

Thus, the percent water is:

\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\

So we plug in to obtain:

\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%

Best regards!

5 0
3 years ago
A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro
Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)

So, OH^{-} is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of OH^{-} increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant.

Therefore excess amount of OH^{-} combines with NH_{4}^{+} to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0
4 years ago
In your own words what is equilibrium?
fenix001 [56]

Answer:

when the forward and reverse reactions occur at equal rates.

chemical reaction is in equilibrium when the concentrations of reactants and products are constant - their ratio does not vary.

8 0
4 years ago
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