As we can see the chemical equation is balanced.K3PO4 + Al(NO3)3 → 3KNO3 + AlPO4
So, by principle of conservation of mass when 1 mole of K3PO4 reacts with 1 mol of Al(NO3)3 , it peoduces 3 mol of KNO3 and 1 mol of AlPO4
So, when 2.5 moles of potassium phosphate react and Al(NO3)3 is present in excess , 2.5*3= 7.5 mol of KNO3 is formed
Answer:
Mole ratio for a compound
The chemical formula tells us the mole ratio.
CO2 = 1 CO2 molecule : 1 C atom : 2 O atoms.
Mole ratio for a reaction
The balanced chemical reactions tells us.
C12H22O11 + 12 O2 12 CO2 + 11 H2O
1 C12H22O11 molecule: 12 O2 molecules : 12 CO2
molecules : 11 H2O molecules.
Applications of the mole ratio concept
grams <--> moles <--> moles <--> grams
Explanation:
Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
Since X is 1 g, therefore O must be 0.1 g. Therefore:
moles O = 0.1 g / (16 g / mol) = 0.00625 mol
We can see that for every 3 moles of O, there are 2 moles
of X, therefore:
moles X = 0.00625 mol O (3 moles X / 2 moles O) =
0.009375 mol
Molar mass X = 1 g / 0.009375 mol
<span>Molar mass X = 106.67 g/mol</span>
