How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?

Chemistry

1 answer:

BartSMP [9]2 years ago

7 0

Answer:

$m_{AgNO_3}=577.6mg$

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:

$n_{LiOH}=81.5mg*\frac{1g}{1000mg}*\frac{1mol}{23.95g} =0.0034molLiOH$

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:

$m_{ AgNO_3}=0.0034molLiOH*\frac{1molAgNO_3}{1molLiOH} *\frac{169.87gAgNO_3}{1molAgNO_3} *\frac{1000gAgNO_3}{1gAgNO_3} \\\\m_{AgNO_3}=577.6mg$

Best regards!

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