Ta có: \(n_{P_2O_5}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
___0,01_____________0,02 (mol)
\(\Rightarrow m_{H_3PO_4}=0,02.98=1,96\left(g\right)\)
Bạn tham khảo nhé!
\(n_{P_2O_5}=\dfrac{1.42}{142}=0.01\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(0.01......................0.02\)
\(m_{H_3PO_4}=0.02\cdot98=1.96\left(g\right)\)