Answer: 4.96 moles
Explanation:
C5H12 is the chemical formula for pentane, the fifth member of the alkane family.
Given that,
number of moles of C5H12 = ?
Mass in grams = 357.4 g
Molar mass of C5H12 = ?
To get the molar mass of C5H12, use the atomic mass of carbon = 12g; and Hydrogen = 1g
i.e C5H12 = (12 x 5) + (1 x 12)
= 60g + 12g
= 72g/mol
Now, apply the formula
Number of moles = Mass / molar mass
Number of moles = 357.4g / 72g/mol
= 4.96 moles
Thus, 4.96 moles of C5H12 that are contained in 357.4 g of the compound.
1 CH4 (g) + 2 O2 (g) -----> CO2 (g) + 2H2O(l) ΔH= - 890 kJ
1 mol 2 mol
1) If ΔH has minus, it means "release". We need only "release" choices.<span>
2) From reaction
1 mol </span>CH4 (g) "releases" ΔH= - 890 kJ - We do not have this choice.
2 mol O2 (g) "release" ΔH= - 890 kJ, so
1 mol O2 (g) "release" ΔH= - 445 kJ
Correct answer is B.
To relate the quantity of a substance to the amount of heat transferred and heat capacity, temperature, number of moles is given by
q=ncpΔT
where
n is number of moles
cp is molar heat capacity
ΔT=Tfinal−Tinitial is the temperature difference
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
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<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
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