If 355 g Al(NO3)3 reacts with an excess of K2CrO4, how many moles of KNO3 would be produced?
2Al(NO3)3 + 3K2CrO4 ==> 6KNO3 + ?
mols Al(NO3)3 = grams/molar mass = ?
Convert to mols KNO3 using the coefficients in the balanced equation.
?mols Al(NO3)2 x [6 mol KNO3/2 mol Al(NO3)3]= ? = mols KNO3.
To determine the number of moles of KNO3 produced, we need to use the balanced chemical equation for the reaction between Al(NO3)3 and K2CrO4.
The balanced equation for the reaction is:
2 Al(NO3)3 + 3 K2CrO4 -> Al2(CrO4)3 + 6 KNO3
From the equation, we see that 2 moles of Al(NO3)3 react with 3 moles of K2CrO4 to produce 6 moles of KNO3.
First, we need to calculate the number of moles of Al(NO3)3 using its given mass. The molar mass of Al(NO3)3 can be calculated as follows:
Al: 26.98 g/mol x 2 = 53.96 g/mol
N: 14.01 g/mol x 6 = 84.06 g/mol
O: 16.00 g/mol x 18 = 288.00 g/mol
Al(NO3)3: 53.96 g/mol + 84.06 g/mol + 288.00 g/mol = 426.02 g/mol
Now, we can find the number of moles of Al(NO3)3:
moles of Al(NO3)3 = mass / molar mass
moles of Al(NO3)3 = 355 g / 426.02 g/mol
Now that we have the number of moles of Al(NO3)3, we can determine the number of moles of KNO3 produced using the stoichiometry of the balanced equation:
moles of KNO3 = (moles of Al(NO3)3 / 2) x 6
Substituting the value we found for moles of Al(NO3)3 into the equation, we can calculate the moles of KNO3 produced.