How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?
1 answer:
Answer:
Explanation:
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In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:
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:
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If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.
<h3>What is Stoichiometry ?</h3>Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.
<h3>What is Balanced Chemical Equation ?</h3>The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now we have to write the balanced equation
Mg + 2HNO₃ → Mg(NO₃)₂ + H₂
According to Stoichiometry
= 3.3 g H₂
Thus from the above conclusion we can say that If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.
Learn more about the Stoichiometry here: brainly.com/question/16060223
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