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=(MW~Q_2P_3)/(2.Ar~Q)* mass Q\\\\mass~Q_2P_3=(102)/(2* 27)* 1.62\\\\mass~Q_2P_3=3.06~g`

Mass P (Ar=16) :

`\tt mass~P=(3.Ar~P)/(MW~Q_2P_3)* mass~Q_2P_3\\\\mass~P=(3.16)/(102)* 3.06\\\\mass~P=1.44~g`

or you can solve it with mol :

Reaction

3P + 2Q ⇒ Q₂P₃

mol Q :

`\tt (1.62)/(27)=0.06`

mol P :

`\tt (3)/(2)* 0.06=0.09`

mass P :

`\tt 0.09* 16=\boxed{\bold{1.44~g}}`