Question

Answer 1

Câu 10:

\(nCaO+mP_2O_5\rightarrow Ca_nP_{2m}O_{5m+n}\)

Có \(\dfrac{31.2m}{40n+31.2m+16\left(5m+n\right)}.100\%=20\%\)

=> \(\dfrac{m}{n}=\dfrac{1}{3}\) => Chọn m = 1; n = 3

CTHH: Ca₃P₂O₈ hay Ca₃(PO₄)₂

PTHH: \(3CaO+P_2O_5\rightarrow Ca_3\left(PO_4\right)_2\)

Câu 11:

\(\left\{{}\begin{matrix}n_{Mg}=2a\left(mol\right)\\n_R=a\left(mol\right)\end{matrix}\right.\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

           2a--------------->2a---->2a

            R + 2HCl --> RCl₂ + H₂

            a--------------->a--->a

=> \(2a+a=\dfrac{6,72}{22,4}=0,3\)

=> a = 0,1 (mol)

Có: m_muối = 0,2.95 + 0,1.(M_R + 71) = 31,7

=> M_R = 56(Fe)