Q3.132CP - Chemistry: Molecular Nature Of Matter And Change

Question

For the reaction between solid tetraphosphorus trisulfide and oxygen gas to form solid tetraphosphorusdecaoxide and sulfur dioxide gas, write a balanced equation. Show the equation (see Table 3.5) in terms grams.

Step-by-Step Solution

Answer

The following reaction we get from the given statement:

${\text{P}}_4{\text{S}}_3\left(\text{s}\right)+8{\text{O}}_2\left(\text{g}\right)\to {\text{P}}_4{\text{O}}_{10}\left(\text{s}\right)+3{\text{SO}}_2\left(\text{g}\right)$

220.093 g of P₄S₃ reacts with 255.984 g O₂ to form 283.886g of P₄O₁₀ and 192.174 g of SO₂.

Step1.Comparison.

The following reaction we get from the given statement:

${\text{P}}_{\text{4}}{\text{S}}_3\left(s\right)+8{\text{O}}_2\left(g\right)\to {\text{P}}_{\text{4}}{\text{O}}_{10}\left(s\right)+3{\text{SO}}_2\left(g\right)$

Molecules:

1 molecule P₄S₃ reacts with 8 molecules O₂ to form 1 molecule P₄O₁₀ and 3 molecules of SO₂

Moles:

1 mole of P₄S₃ reacts with 8 moles O₂ to form 1 mole P₄O₁₀ and 1 moles of SO₂ .

Mass:

$$\begin{gathered} 1{\text{ mol of P}}_4{\text{S}}_3=4(30.974)+3(32.06)=220.093 \\ 1\text{ mol }{O}_2=31.998\text{ g} \\ 1{\text{ mol P}}_{\text{4}}{\text{O}}_{10}=283.886\text{g} \\ 1{\text{ mol SO}}_2=64.058\text{g} \end{gathered}$$

Step2.Final Result.

Multiply the coefficient of 8 with 1 mol O₂ and others with their respective coefficient, now we get:

Mass:

$$\begin{gathered} 1{\text{ mol of P}}_4{\text{S}}_3=4(30.974)+3(32.06)=220.093 \\ 1{\text{ mol O}}_2=31.998\text{ g}\times 8=255.984 \\ 1{\text{ mol P}}_4{\text{O}}_{10}=283.886 \text{g} \\ 1{\text{ mol SO}}_2=64.058 \text{g}\times 3=192.174 \end{gathered}$$

220.093 g of P₄S₃ reacts with 255.984 g O₂ to form 283.886g of P₄O₁₀ and 192.174 g of SO₂.