When 1.0 mole of Cu2FeS3 and 1.0 mole of O2 are mixed and allowed to react according to the equation
$$2Cu_2FeS_3(s) + 7O_2(g) \rightarrow 6Cu(s) + 2Fe_2O_3(s) + 6SO_2(g)$$
A - VCE - SSCE Chemistry - Question 4 - 2010 - Paper 1
Question 4
When 1.0 mole of Cu2FeS3 and 1.0 mole of O2 are mixed and allowed to react according to the equation
$$2Cu_2FeS_3(s) + 7O_2(g) \rightarrow 6Cu(s) + 2Fe_2O_3(s) + 6S... show full transcript
Worked Solution & Example Answer:When 1.0 mole of Cu2FeS3 and 1.0 mole of O2 are mixed and allowed to react according to the equation
$$2Cu_2FeS_3(s) + 7O_2(g) \rightarrow 6Cu(s) + 2Fe_2O_3(s) + 6SO_2(g)$$
A - VCE - SSCE Chemistry - Question 4 - 2010 - Paper 1
Step 1
Calculate the moles of each reactant.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Given 1.0 mole of Cu2FeS3 and 1.0 mole of O2. The balanced reaction shows that 2 moles of Cu2FeS3 react with 7 moles of O2.
Step 2
Determine the stoichiometric ratios.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
From the equation: (2) moles of Cu2FeS3 need (7) moles of O2. Therefore, for 1 mole of Cu2FeS3, the required O2 is (\frac{7}{2} = 3.5) moles. Since we only have 1 mole of O2, Cu2FeS3 is in excess.
Step 3
Calculate the excess moles of Cu2FeS3.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Since we have only 1 mole of O2, we can find how much Cu2FeS3 is used. (1 , mole , O2) will react with (\frac{2}{7} , moles , Cu2FeS3). Therefore, the moles of Cu2FeS3 left unreacted:
[ 1.0 - \frac{2}{7} = \frac{5}{7} , moles , Cu2FeS3 , excess ]
Step 4
Select the correct option.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Based on the calculations, (\frac{5}{7}) mole of Cu2FeS3 is in excess (Option C).
