One mole methane is completly burned in xcess oxygen. If 8.90x10^3kj of heat are given off from theis reaction, calculate the heat given off for 1.44g of methane.
CH4 + 2O2 ==> CO2 + 2H2O
8.90 x 10^3 kJ given off for 1 mole(16 g) CH4.
1.44 g will give off
8.90 x 10^3 kJ x (1.44/16) = ??
who the hell cares
To calculate the heat given off for 1.44g of methane, we need to use the concept of mole ratio. First, we need to find the number of moles of methane in 1.44g.
Molar mass of methane (CH4) = 12.01g/mol (carbon) + (4 * 1.01g/mol) (hydrogen) = 16.05g/mol
Number of moles of methane = (mass of methane / molar mass of methane) = (1.44g / 16.05g/mol)
Next, we can use the molar ratio between methane and heat given off from the balanced chemical equation. The balanced equation for the complete combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O
From the balanced equation, we can see that the mole ratio between methane and heat is 1:1. In other words, one mole of methane gives off a certain amount of heat, which is 8.90x10^3 kJ.
Finally, we can calculate the heat given off for 1.44g of methane:
Heat given off = (Number of moles of methane) * (Heat given off per mole)
Heat given off = (1.44g / 16.05g/mol) * (8.90x10^3 kJ/mol)
Solving this equation will give us the heat given off for 1.44g of methane.