In a spontaneous oxidation-reduction reaction between aluminum and silver ion, the aluminum(III) ion and solid silver are formed. If 0.001g of aluminum is consumed in this reaction, how much silver is produced?
a)1.20g
b)1.00g
c)3.00g
d)530g
Here is an example stoichiometry problem I've posted. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the amount of silver produced in the reaction, we can use stoichiometry to relate the amount of aluminum consumed to the amount of silver produced.
First, we need to balance the chemical equation for the reaction:
2 Al + 3 Ag⁺ → 2 Al³⁺ + 3 Ag
From the balanced equation, we can see that 2 moles of aluminum react to form 3 moles of silver.
Next, we need to convert the given mass of aluminum (0.001g) to moles. To do this, we use the molar mass of aluminum, which is 26.98 g/mol:
moles of aluminum = mass of aluminum / molar mass of aluminum
moles of aluminum = 0.001g / 26.98 g/mol
moles of aluminum = 0.000037 mol
Now that we have the number of moles of aluminum, we can use the stoichiometry of the balanced equation to find the number of moles of silver produced.
From the balanced equation, we know that:
2 moles of aluminum react to produce 3 moles of silver
Therefore, using the ratio of moles of aluminum to moles of silver, we can calculate the moles of silver produced:
moles of silver = (moles of aluminum) x (3 moles of silver / 2 moles of aluminum)
moles of silver = 0.000037 mol x (3 mol / 2 mol)
moles of silver = 0.0000555 mol
Finally, we convert the moles of silver to grams using the molar mass of silver, which is 107.87 g/mol:
mass of silver = moles of silver x molar mass of silver
mass of silver = 0.0000555 mol x 107.87 g/mol
mass of silver = 0.00599g
Therefore, the amount of silver produced in the reaction is approximately 0.006g.
Since none of the answer choices seem to match the calculated value of 0.006g, it is likely that there is an error in the options provided for the question.