First we convert the given reactant masses into moles, using their respective molar masses:
4.00 g H₂ ÷ 2 g/mol = 2 mol H₂
6.20 g P₄ ÷ 124 g/mol = 0.05 mol P₄
0.05 moles of P₄ would react completely with (6*0.05) 0.3 moles of H₂. There are more H₂ moles than required, meaning H₂ is in excess and P₄ is the limiting reactant.
Now we<u> calculate how many PH₃ moles could be formed</u>, using the <em>number of moles of the limiting reactant</em>:
0.05 mol P₄ * = 0.2 mol PH₃
Finally we <u>convert 0.2 mol PH₃ into grams</u>, using its <em>molar mass</em>:
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