Question

Cho 7,1 gam P₂O₅ hòa tan hết trong 100 gam nước. Tính nồng độ phần trăm dung dịch thu được

 

Answer 1

\(n_{P_2O_5}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

\(P_2O_5+3H_2O\xrightarrow[]{}2H_3PO_4\)

0,05 → 0,15    →     0,1

\(\Rightarrow m_{H_3PO_4}=0,1\cdot98=9,8\left(g\right)\)

\(\Rightarrow m_{H_2O}\left(\text{pư}\right)=0,15\cdot18=2,7\left(g\right)\)

\(\Rightarrow m_{H_2O}\left(\text{dm}\right)=100-2,7=97,3\left(g\right)\)

\(\Rightarrow m_{H_3PO_4}\left(\text{dd}\right)=m_{H_3PO_4}+m_{H_2O}\left(\text{dm}\right)=9,8+97,3=107,1\left(g\right)\)

\(\Rightarrow C\%=\dfrac{m_{H_3PO_4}}{m_{H_3PO_4}\left(\text{dd}\right)}\cdot100\%=\dfrac{9,8}{107,1}\cdot100\%\approx9,15\%\)