the required energy is... 55.5 multiply by 286 = 15873 K.J....
You have 42.5 g of water. 42.5g H20 / 18.02 g H2O (2.358 moles H2O)*(6.02 Kj/1mole H2O) = 14.1981 Kj =14.2 kJ
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
3,75 moles hydrogen
Pure combustion of butane has the reaction 2 C4H10 + 13 O2 equals8 CO2 + 10 H2O. This means that every mole of butane produces 5 moles of water. 20 ml of water equals 1.11 moles, so .222 moles of butane are needed, or 5.17 liters.
50mL minus the volume of 0.3 moles pure HCl is how much water is necessary. Be sure to add acid to water never water to acid.
The necessary heat is 9,22 joules.
yes moles do drin water but not much
72.8grams
Multiply moles by molecular mass of water (18), gives you 223.8g. Remember this formula: Number of moles = mass / molecular mass
Yes,but not too much...
The energy is 18,263.10e4 joules.
You have 42.5 g of water. 42.5g H20 / 18.02 g H2O (2.358 moles H2O)*(6.02 Kj/1mole H2O) = 14.1981 Kj =14.2 kJ
2H2 + O2 ---------------> 2H2O for every 2 moles of hydrogen that reacts, 2 moles of water are produced, thus a 1:1 ratio of water produced to hydrogen reacted. So:- 2.5 moles of hydrogen reacted will produce 2.5 moles of water
548kJ
540 calories per gram is absorbed when water vaporizes at its boiling point. Called the latent heat of vaporization. 540 x 23.1 x 18 = 224532 calories
Calculate the mass in grams of water vapor produced if 3.11 moles of propane is burned
As the molar mass of water is 18 g/mol (1*2 H + 16 O) 10 moles*18 g/mole=180 grams.