Question

what is the mass of element P which reacts with 1 .62g of element Q to form a compound with formula Q2p3
(relative atomic mass p,16;Q,27)​

Answer 1

Mass of element P = 1.44 g

Further explanation

Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements

In the same compound, although from different sources and formed by different processes, it will still have the same composition/comparison

MW Q₂P₃ = 2.27 + 3.16=102 g/mol

Mass Q₂P₃ (MW=102 g/mol) :

$\tt mass~Q_2P_3=\dfrac{MW~Q_2P_3}{2.Ar~Q}\times mass Q\\\\mass~Q_2P_3=\dfrac{102}{2\times 27}\times 1.62\\\\mass~Q_2P_3=3.06~g$

Mass P (Ar=16) :

$\tt mass~P=\dfrac{3.Ar~P}{MW~Q_2P_3}\times mass~Q_2P_3\\\\mass~P=\dfrac{3.16}{102}\times 3.06\\\\mass~P=1.44~g$

or you can solve it with mol :

Reaction

3P + 2Q ⇒ Q₂P₃

mol Q :

$\tt \dfrac{1.62}{27}=0.06$

mol P :

$\tt \dfrac{3}{2}\times 0.06=0.09$

mass P :

$\tt 0.09\times 16=\boxed{\bold{1.44~g}}$