0.224g sample that contained only bacl2 and kbr required 19.7ml of 0.1M agn03 to react the end point. Calculate the percent of each compound present in the sample
To calculate the percent of each compound present in the sample, we need to use stoichiometry and the given information.
Step 1: Determine the moles of AgNO3 used.
The volume (V) of AgNO3 used is given as 19.7 mL, and the concentration (C) of AgNO3 is 0.1 M.
Using the formula C = n/V, we can rearrange it to find the moles (n) of AgNO3:
n = C * V
n = 0.1 mol/L * 0.0197 L
n = 0.00197 mol
Step 2: Determine the moles of BaCl2 and KBr based on the reaction stoichiometry.
From the balanced equation:
2 AgNO3 + BaCl2 -> 2 AgCl + Ba(NO3)2
1 AgNO3 + KBr -> AgBr + KNO3
We can see that each mole of AgNO3 reacts with 1 mole of BaCl2 and 1 mole of KBr.
Since we have measured the moles of AgNO3 used (0.00197 mol), the amount of both BaCl2 and KBr would be the same.
Step 3: Determine the mass of BaCl2 and KBr.
The molar mass of BaCl2 is 137.33 g/mol, and the molar mass of KBr is 119 g/mol.
Using the formula mass = moles * molar mass, we can calculate the mass of each compound:
mass of BaCl2 = 0.00197 mol * 137.33 g/mol
mass of BaCl2 = 0.270 g
mass of KBr = 0.00197 mol * 119 g/mol
mass of KBr = 0.235 g
Step 4: Calculate the percent of each compound.
The percent of each compound is calculated by dividing the mass of the compound by the total mass (0.224 g) and multiplying by 100%:
percent of BaCl2 = (mass of BaCl2 / total mass) * 100%
percent of BaCl2 = (0.270 g / 0.224 g) * 100%
percent of BaCl2 = 120.54%
percent of KBr = (mass of KBr / total mass) * 100%
percent of KBr = (0.235 g / 0.224 g) * 100%
percent of KBr = 104.91%
Therefore, the percent of BaCl2 in the sample is approximately 120.54% and the percent of KBr in the sample is approximately 104.91%.
Note: It seems that there might be an error in the given data as the percentages exceed 100%. Please double-check your data or calculation if these results seem unexpected.