1) If 570 mL of 0.96491 M aqueous CaCl 2 reacts stoichiometrically according to
ID: 498773 • Letter: 1
Question
1) If 570 mL of 0.96491 M aqueous CaCl2 reacts stoichiometrically according to the balanced equation, how many grams of solid Ca3(PO4)2 are produced?
3CaCl2(aq) + 2Na3PO4(aq) Ca3(PO4)2(s) + 6NaCl(aq)
2) If 490 mL of 1.73 M aqueous H2SO4 and 0.810 mol of solid CaCO3 are reacted stoichiometrically according to the balanced equation, how many milliliters of 1.73 M aqueous H2SO4 remain? Round your answer to 3 significant figures.
H2SO4(aq) + CaCO3(s) CO2(g) + CaSO4(s) + H2O(l)
3) If 0.390 mol of gaseous C2H4 reacts stoichiometrically according to the balanced equation, how many liters of gaseous C2H4O measured at STP are produced?
2C2H4(g) + O2(g) 2C2H4O(g)
Explanation / Answer
Answer for question (1):
Reaction Eqation : 3CaCl2(aq) + 2Na3PO4(aq) Ca3(PO4)2(s) + 6NaCl(aq)
Molarity (M) of CaCl2 solution used: 0.96491 M
That is to say 1 ml of 0.96491 M contains 0.96491 milli moles in it.
Volume (V) of CaCl2 solution used in reaction: 570 ml
Therefore total number of milli moles of CaCl2 used in reaction: V x M = 570 x 0.96491 = 550 mmoles
Weight of CaCl2 used: no. of mmoles x Mol wt/1000 = 550 x 110.98/1000 = 61.039 g
According to the stoichiometric equation 3 mmoles of CaCl2 gives 1 mmole of Tricalcium phosphate
CaCl2 Molecular wt = 110.98 g/mol
= 0.110 g/mmol
Ca3(PO4)2 Molecular wt = 310.18 g/mol
= 0.310 g/mmol
For preparing 1 mmol (0.310 g) of Ca3(PO4)2 , 3 mmol (0.110x3 0.330 g) of CaCl2 is required i.e to say
3 mmol CaCl2 gives 1 mmol Ca3(PO4)2 and so in the above solution if 550 mmole CaCl2 is used.
Then 550/3 = 183.333 mmoles of Ca3(PO4)2 is obtained.
Accordingly 183.333 x 0.310 g = 56.833 g of Ca3(PO4)2 obtained
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Answer for question (2):
Reaction: H2SO4(aq) + CaCO3(s) CO2(g) + CaSO4(s) + H2O(l)
Stoichometrically for 1 mmol of H2SO4 1 mmol of CaCO3 is require for completion of reaction.
According to question (2) 490 ml of 1.73 M H2SO4 is used.
Therefore total number of mmoles of H2SO4 used = 490 x 1.73
= 847.7 mmoles
Total number of mmoles of CaCO3 used = 810 mmoles
Excess number of mmoles of H2SO4 used = Total number of mmoles of H2SO4 used - Total number of mmoles of CaCO3 used
Therefore excess number of mmoles of H2SO4 used = 847.7 - 810
= 37.7 mmoles
The H2SO4 concentration is 1.73 M, i.e 1 ml of it contains 1.73 mmoles of H2SO4
Therefore 37.7 mmoles of H2SO4 will be in = 37.7/1.73 = 22.176 ml of 1.73 M solution of H2SO4
So 22.176 ml of 1.73 M solution of H2SO4 is remained in excess
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Answer for question (3):
Reaction equation: 2C2H4(g) + O2(g) 2C2H4O(g)
Stoichiometrically according to the balanced equation 2 moles of ethane gas upon reaction with 1 mole of oxygen gas gives 2 moles of gaseous C2H4O
Ethane gas used in the question = 0.39 M
So according to the balanced equation 0.39 moles of ethane should give 0.39 M of gaseous C2H4O.
At STP one mole of gas = 22.4 L
Therefore 0.39 M x 22.4 L = 8.736 L of C2H4O are produced
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