3Ag(s) + 4HNO3(AQ) = 3AGNO3(aq) + NO(g) + 2h2o(l)
How many grams of AgNO3 can be produced by recating completely 88.0g of Ag?
please show me the step.. thank you
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To determine how many grams of AgNO3 can be produced by reacting 88.0g of Ag, you need to calculate the stoichiometry of the reaction.
1. Begin by writing down the balanced chemical equation given in the question:
3Ag(s) + 4HNO3(aq) → 3AgNO3(aq) + NO(g) + 2H2O(l)
2. Identify the stoichiometric ratio between Ag and AgNO3. In the balanced equation, it shows that 3 moles of Ag produce 3 moles of AgNO3.
3. Calculate the molar mass of AgNO3. The molar mass of AgNO3 is the sum of the atomic masses of Ag, N, and three O atoms:
AgNO3: (Ag) + (N) + 3(O) = 107.87 g/mol + 14.01 g/mol + (16.00 g/mol x 3) = 169.87 g/mol
4. Convert grams of Ag to moles. Divide the given mass of Ag by its molar mass:
88.0g Ag ÷ 107.87 g/mol = 0.815 mol Ag (rounded to three decimal places)
5. Use the stoichiometric ratio to determine the moles of AgNO3 produced. Since the stoichiometric ratio is 1:1 between Ag and AgNO3, the number of moles of AgNO3 produced will also be 0.815 mol.
6. Convert moles of AgNO3 to grams. Multiply the moles of AgNO3 by its molar mass:
0.815 mol AgNO3 x 169.87 g/mol = 138.5 g AgNO3 (rounded to one decimal place)
Therefore, when 88.0g of Ag reacts completely, it will produce approximately 138.5g of AgNO3.