Question

how many grams of phosphorus (p₄) are needed to completely consume 46.1 l of chlorine gas according to the following reaction at 25°c and 1 atm? phosphorus (p₄) (s) + chlorine (g) → phosphorus trichloride (l) grams phosphorus (p₄) use the references to access important values if needed for this question. 9 more group attempts remaining

Explanation:

Step1: Calculate moles of chlorine gas

Use the ideal gas law \(PV = nRT\). Rearranged for \(n\): \(n=\frac{PV}{RT}\).
Given \(P = 1\ atm\), \(V=46.1\ L\), \(R = 0.0821\ L\cdot atm/(mol\cdot K)\), \(T=(25 + 273)K=298\ K\).
\(n=\frac{1\ atm\times46.1\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times298\ K}\)
\(n=\frac{46.1}{24.4658}\ mol\approx1.884\ mol\)

Step2: Use stoichiometry from reaction

The balanced reaction for \(P_4+6Cl_2
ightarrow4PCl_3\) (assuming the reaction to form phosphorus trichloride).
From the stoichiometry, \(n(P_4)=\frac{n(Cl_2)}{6}\)
Since \(n(Cl_2) = 1.884\ mol\), \(n(P_4)=\frac{1.884}{6}\ mol = 0.314\ mol\)

Step3: Calculate mass of \(P_4\)

Molar mass of \(P_4\) is \(M = 4\times30.97\ g/mol=123.88\ g/mol\)
Mass \(m=n\times M\)
\(m = 0.314\ mol\times123.88\ g/mol\approx38.9\ g\)

Answer:

\(38.9\ g\)