a)
\(n_{AgNO_3}=0,15.2=0,3\left(mol\right)\)
\(2AgNO_3+BaCl_2\rightarrow2AgCl\downarrow+Ba\left(NO_3\right)_2\)
0,3-------->0,15-------------------->0,15
b)
\(V_{dd.BaCl_2}=\dfrac{0,15}{2}=0,075\left(l\right)\\ V_{dd.sau.pư}=0,075+0,15=0,225\left(l\right)\\ C_{M\left(Ba\left(NO_3\right)_2\right)}=\dfrac{0,15}{0,225}=\dfrac{2}{3}M\)
mình làm lại nha cái kia làm sai rồi
\(V_{AgNO_3}=150ml=0,15\left(l\right)\)
=> \(n_{AgNO_3}=CM.V=2.0,15=0,3\left(mol\right)\)
\(2AgNO_3+BaCl_2\rightarrow2AgCl+Ba\left(NO_3\right)_2\)
0,3 0,15 0,3 0,15 (mol)
\(V_{BaCl_2}=\dfrac{n}{CM}=\dfrac{0,15}{2}=0,075\left(l\right)\)
\(V_{Ba\left(NO_3\right)_2}=0,15+0,075=0,225\left(mol\right)\)
=> \(CM_{Ba\left(NO_3\right)_2}=\dfrac{n}{V}=\dfrac{0,15}{0,225}=\dfrac{2}{3}M\)
