\(m_{Na_3PO_4}=\frac{328.15}{100}=49,2\left(g\right)\\ \rightarrow n_{Na_3PO_4}=\frac{m}{M}=\frac{49,2}{164}=0,3\left(mol\right)\)
\(PTHH:2Na_3PO_4+3CaCl_2\rightarrow Ca_3\left(PO_4\right)_2+6NaCl\)
(mol) 2 3 1 6
(mol) 0,3 0,45 0,15 0,9
\(m_{CaCl_2}=n.M=0,45.111=49,95\left(g\right)\)
\(m_{Ca_3\left(PO_4\right)_2}=n.M=0,15.310=46,5\left(g\right)\)
\(m_{NaCl}=n.M=58,5.0,9=52,65\left(g\right)\)
\(C\%_{ddM_1}=\frac{46,5}{328+49,95-52,65}.100\%=12,29\left(\%\right)\)
\(C\%_{ddM_{NaCl}}=\frac{52,65}{328+49,95-46,5}.100\%=15,88\left(\%\right)\)
Câu C% mình làm sai nhé! Theo bài chị PHC ấy!
\(V_{Na_3PO_4}=\frac{328}{1,3}=252,3\left(ml\right)=0,252\left(l\right)\\ V_{CaCl_2}=\frac{\left(\frac{49,95.100}{20}\right)}{1,12}=223\left(ml\right)=0,223\left(l\right)\)
\(C_{M_{Ca_3PO_4}}=\frac{0,15}{0,252+0,223}=0,3\left(M\right)\)
\(C_{M_{NaCl}}=\frac{0,9}{0,252+0,223}=1,9\left(M\right)\)