Someone please help me pass gen chem Suppose a boil water notice is sent out advising all residents in the area to boil their water before drinking it or using it for cooking. You need to boil 16.5 L of water using your natural gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 17.9% of the heat generated goes towards heating the water. Assume the density of methane is 0.668 g/L, the density of water is 1.00 g/mL, and that the water has an initial temperature of 20.4 °C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane.

The volume of natural gas needed to boil the water if only 17.9% of the heat is generated towards heating water is ; 918.70 L

Using the given data :

Volume of water = 16.5 L = 16500 mL

mass of water = 16500 g  ( 16500 mL * 1.00 g/mL )

Density of methane = 0.668 g/L

Density of water = 1.00 g/mL

First step : determine the heat needed to boil the water

Heat required = mass of water * specific heat water * ( Δ T )

                        = 16500 * 4.184 * ( 100 - 20.4 ) =  5495265.6 J

                        = 5495.265 kJ

∴ Heat required to boil water = 5495.265 kJ

next step ; determine the heat given by combustion

heat given by combustion = ( Heat required to boil water) / ( % of heat generated )

Heat given by combustion = ( 5495.265 ) / ( 17.9 % )

                                             = 30699.80 kJ

Enthalpy of methane combustion = 802.5 kJ/mol

∴ moles of methane used = ( 30699.80 ) / ( 802.5 ) = 38.26 mol

next ; determine the mass of methane ( natural gas )

= ( moles of methane used ) * ( molar mass )

= 38.26 * 16.04 g/mol = 613.69 g

Final step : Calculate the volume of natural gas is needed to boil the water

= mass of natural gas / density of methane

= 613.69 g / 0.668 g/L

= 918.70 L

Hence we can conclude that the volume of natural gas needed to boil water if only 17.9% of the heat is  918.70 L .

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Solution-

From the question the volume of water = 18 L = 18000 mL

Now we can find the mass of water = (volume of water) * (density of water)

mass of water = (18000 mL) * (1.00 g/mL)

mass of water = 18000 g

Now we find the heat required to boil water = (mass of water) * (specific heat water) * (final temperature - initial temperature)

putting the value the heat required to boil water = (18000 g) * (4.184 J/g.oC) * (100 oC - 22.7 oC)

heat required to boil water = 5821617.6 J

heat required to boil water = 5821.62 kJ

The heat given by the combustion = (heat required to boil water) / (percent of heat taken by boiling)

Heat given by combustion = (5821.62 kJ) / (19.4 /100)

Therefore the heat given by combustion = (5821.62 kJ) / (0.194)

Heat given by combustion = 30008.35 kJ

As we know that the enthalpy of combustion of methane = 802.5 kJ/mol

The moles methane used = (Heat given by combustion) / (enthalpy of combustion of methane)

moles methane used = (30008.35 kJ) / (802.5 kJ/mol)

So the moles methane used = 37.39 mol

Now the mass methane = (moles methane used) * (molar mass methane)

The mass methane = (37.39 mol) * (16.04 g/mol)

The mass methane = 599.74 g

Now the volume methane = (mass methane) / (density of methane)

volume methane = (599.7356 g) / (0.660 g/L)

volume methane = 908.69 L

hope helped!!

plz mark brainliest:DD

the answer would be c