Answer:
[tex]m_{AgNO_3}=577.6mg[/tex]
Explanation:
Hello there!
In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:
[tex]n_{LiOH}=81.5mg*\frac{1g}{1000mg}*\frac{1mol}{23.95g} =0.0034molLiOH[/tex]
Thus, since there is a 1:1 mole ratio between lithium hydroxide and silver nitrate (169.87 g/mol) the resulting milligrams turn out to be:
[tex]m_{ AgNO_3}=0.0034molLiOH*\frac{1molAgNO_3}{1molLiOH} *\frac{169.87gAgNO_3}{1molAgNO_3} *\frac{1000gAgNO_3}{1gAgNO_3} \\\\m_{AgNO_3}=577.6mg[/tex]
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