Answer:
[tex]m_{NaCl}=13.2gNaCl[/tex]
Explanation:
Hello,
In this case, since the reaction is:
NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
And it is properly balanced, given 1.50 L of 0.150-M AgNO3, the first step is to compute the moles of reacting silver nitrate as follows given its volume and concentration:
[tex]n_{AgNO_3}=1.50L*0.150mol/L\\\\n_{AgNO_3}=0.225mol[/tex]
Next, by applying stoihiometry, given the 1:1 mole ratio between silver nitrate and sodium chloride, we compute the mass of NaCl by also using its molar mass (58.45 g/mol):
[tex]m_{NaCl}=0.225molAgNO_3*\frac{1molNaCl}{1molAgNO_3}*\frac{58.45gNaCl}{1molNaCl}\\ \\m_{NaCl}=13.2gNaCl[/tex]
Regards.
