Question

The compound As2I4 is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic (d = 5.72 g/cm3 ) that is 3.00 cm on edge is allowed to react with
1.01 * 10^24 molecules of arsenic triiodide, how much As2I4 can be prepared? If the percent yield of As2I4 was 75.6%, what mass of As2I4 was actually isolated?

Answer 1

Mass As₂I₄ actually isolated : 626.27 g

Further explanation

Reaction

2As+4AsI₃⇒3As₂I₄

mass As : ρ x V

$\tt 5.72\times 3^3=5.72\times 27=154.44~g$

mol As (MW 75 g/mol) :

$\tt \dfrac{154.44}{75}=2.059$

mol AsI₃ :

$\tt \dfrac{1.01\times 10^{24}}{6.02\times 10^{23}}=1.68$

Limiting reactant :

As:AsI₃ =

$\tt \dfrac{2.059}{2}:\dfrac{1.68}{4}=1.03:0.42(AsI_3~smaller\rightarrow limiting~reactant)$

mol As₂I₄ :

$\tt \dfrac{3}{4}\times 1.68=1.26$

mass As₂I₄ (Mw=657.461 g/mol)

$\tt 1.26\times 657.461=828.4~g$

For 75.6% yield :

$\tt 0.756\times 828.4=626.27~g$