How many grams of methane must be combusted to heat 1.05 of water from 24.0 to 89.0, assuming as a product and 100% efficiency in heat transfer?

from 24.0 to 89.0, assuming .........as a product and........

You omitted something. And do you know the heat of combustion in kJ/mol or kJ/g?

To determine the number of grams of methane required, we need to calculate the amount of heat needed to raise the temperature of the water using the formula:

Q = mcΔT

Where:
Q is the amount of heat energy required (in Joules)
m is the mass of water (in grams)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (89.0°C - 24.0°C)

First, let's calculate Q:

Q = (m)(4.18 J/g°C)(ΔT)

To find the mass of methane required, we will use the balanced chemical equation for the combustion of methane:

CH4 + 2O2 → CO2 + 2H2O

From the equation, we can see that 1 mole of methane (CH4) produces 2 moles of water (H2O) when fully combusted.

To calculate the number of moles of water produced, we'll divide the number of moles by the molar mass of water (H2O) which is 18.015 g/mol.

1.05 moles of water × (1 mole CH4 / 2 moles H2O) × (16.04 g CH4 / 1 mole CH4) = X grams of methane

So, by plugging the values into the equation, we have:

Q = (X g)(4.18 J/g°C)(89.0°C - 24.0°C)

Now we can solve for X by rearranging the equation to find the mass of methane:

X g = Q / [(4.18 J/g°C)(89.0°C - 24.0°C)]

Finally, we substitute the values into the equation to find X:

X g = Q / (4.18 J/g°C × 65.0°C)

Note: The value for Q must be given in Joules to ensure consistent units in the calculation.

Once X is calculated, it will give you the mass of methane required in grams.