Question

Natural gas burns in air to form carbon dioxide and water, releasing heat. CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔHrxn = -802.3 kJ.
What minimum mass of CH4 is required to heat 45.0 g of water by 21.0 ∘C? (Assume 100% heating efficiency.) (For water,Cs= 4.18 J/g∘C).
What minimum mass of is required to heat 50.0 of water by 26.0 ?

Answer 1

Answer:

1) Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

Explanation:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g) ,\Delta H_{rxn} =-802.3 kJ$

1) Minimum mass of  methane required to raise the temperature of water by 21.0°C.

Mass of water = m = 45.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 21.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 45.0 g\times 4.18 J/g^oC\times 21.0^oC$

Q = 3,950.1 J = 3.9501 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 3.950.1 kJ of heat will be given by:

$=\frac{3.950.1 kJ}{802.3 kJ}=0.004923 mol$

Mass of 0.004923 moles of methane :

0.004923 mol × 16 g/mol=0.0788 g

Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

2) Minimum mass of  methane required to raise the temperature of water by 26.0°C.

Mass of water = m = 50.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 26.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

$Q=mc\Delta T= 50.0 g\times 4.18 J/g^oC\times 26.0^oC$

Q = 5,434 J= 5.434 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 5.434 kJ of heat will be given by:

$=\frac{5.434 kJ}{802.3 kJ}=0.006773 mol$

Mass of 0.006773 moles of methane :

0.006773 mol × 16 g/mol= 0.108 g

Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

Answer 2

The minimum mass of methane required to heat up 45g by 21°C is 0.079g

Data;

• Mass = 45g
• Temperature = 21°C

Mass of Methane Needed

From the chemical equation, we can say that each gram of water requires 4.18g of heat to increase by 1°C

$45* 21 = 945J$

945J of heat is required to raise 45g of methane

One mole of methane weighs 16.04g/mol and burning one mole supplied 802.3kJ of heat.

The specific heat Q is given as

$Q = _mC_pdt$

The heat added is the product of the specific heat, the mass and the change in temperature.

The heat added is

$\delta H=4.18*45*(21-0)\\\delta H = 3950J = 3.950kJ$

The number of moles of methane needed is

$\frac{\delta H}{\delta H_r} = \frac{3.950}{802.3} = 4.9234*10^-^3 mols$

The mass of methane needed will be

$mass = 4.9234*10^-^3 * 16.04 = 0.079g$

The minimum mass of methane required to heat up 45g by 21°C is 0.079g

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