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Alexxandr [17]
3 years ago
13

What mass of sulfur must be used to produce 17.5 L of gaseous sulfur dioxide at STP according to the following equation?

Chemistry
1 answer:
Greeley [361]3 years ago
3 0

Answer:

                     25.05 g of S₈

Explanation:

                     The balance chemical equation for given synthetic reaction is,

                                            S₈ + 8 O₂ → 8 SO₂

Step 1: <u>Calculate Moles of SO₂ at STP:</u>

According to Avogadro's Law, at STP (0 ° C and 1 atm) every ideal gas occupies 22.4 L of volume.  Therefore, the moles contained by 17.5 L of SO₂ are as;

                             Moles  =  17.5 L / 22.4 L

                             Moles  =  0.781 moles of SO₂

Step 2: <u>Calculate Moles of S₈ consumed:</u>

According to balance equation,

8 moles of SO₂ were produced by  =  1 mole of S₈

So,

0.781 moles of SO₂ will be produced by  =  X moles of S₈

Solving for X,

                      X  =  0.781 mol × 1 mol / 8 mol

                      X  =  0.0976 moles of S₈

Step 3: <u>Calculate Mass of S₈ as;</u>

                      Mass  =  Moles × M.Mass

                      Mass  =  0.0976 mol × 256.52 g/mol

                      Mass =  25.05 g of S₈

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The formula for the calculation of moles is shown below:

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Moles= \frac{0.236\ g}{233.43\ g/mol}

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Explanation:

It is given that,

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The amount of energy change during the transition is given by :

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Plugging all the values we get :

\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5

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