For the following chemical reaction, how many moles of silver iodide (AgI) will be produced from 4 mol of calcium iodide (CaI2)?
You don't give a reaction; I assume you just didn't type it into the post. I'll write one.
CaI2 + 2AgNO3 ==> 2AgI + Ca(NO3)2
4 mols CaI2.
Using the coefficients in the balanced equation, convert mols CaI2 to mols AgI.
The balanced chemical equation for the reaction between calcium iodide (CaI2) and silver nitrate (AgNO3) to form silver iodide (AgI) and calcium nitrate (Ca(NO3)2) is:
CaI2 + 2AgNO3 -> 2AgI + Ca(NO3)2
From the equation, we can see that 1 mole of calcium iodide (CaI2) reacts with 2 moles of silver nitrate (AgNO3) to produce 2 moles of silver iodide (AgI).
Since there is a 2:1 ratio between calcium iodide and silver iodide, for every 2 moles of calcium iodide used, 2 moles of silver iodide are produced.
Therefore, from 4 moles of calcium iodide (CaI2), we can expect to produce 4 moles of silver iodide (AgI).
To determine the number of moles of silver iodide (AgI) produced from 4 mol of calcium iodide (CaI2), we need to examine the balanced chemical equation for the reaction. Assuming that the reaction takes place between calcium iodide and silver nitrate (AgNO3), the balanced equation is as follows:
CaI2 + 2AgNO3 -> 2AgI + Ca(NO3)2
From the balanced equation, we can see that one mole of calcium iodide (CaI2) reacts with 2 moles of silver nitrate (AgNO3) to produce 2 moles of silver iodide (AgI).
Now, we can use this information to calculate the number of moles of AgI produced from 4 moles of CaI2:
4 mol CaI2 * (2 mol AgI / 1 mol CaI2) = 8 mol AgI
Therefore, 4 moles of calcium iodide will produce 8 moles of silver iodide.