Answer:
67.4 % of C₉H₈O₄
Explanation:
To make titrations problems we know, that in the endpoint:
mmoles of acid = mmoles of base
mmoles = M . volume so:
mmoles of acid = 20.52 mL . 0.1121 M
mmoles of acid = mg of acid / PM (mg /mmoles)
Let's determine the PM of aspirin:
12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol
mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol
mass (mg) = 414.4 mg
We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g
We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %
Answer: Thus concentration of 
in 
is 0.011 and in 
is 0.814
Explanation:
To calculate the concentration of , we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
The concentration in is 
Thus concentration of is 
and 
Answer:
2.05moles
Explanation:
The balanced chemical equation in this question is as follows;
Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O
Based on the above equation, 2 moles of H2SO4 reacted to produce 1 mole of SnSO4
However, the mass of SnSO4 produced is 219.65 grams. Using mole = mass/molar mass, we can find the number of moles of SnSO4 produced.
Molar mass of SnSO4 where Sn = 118.7, S = 32, O = 16
= 118.7 + 32 + 16(4)
= 150.7 + 64
= 214.7g/mol
mole = 219.65/214.7
mole = 1.023mol
Therefore, if 2 moles of H2SO4 reacted to produce 1 mole of SnSO4
1.023 mol of SnSO4 produced will cause: 1.023 × 2/1
= 2.046moles of H2SO4 to react.
