During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

What is the amount of the leftover reactant?

0.74 grams of methane

0.89 grams of methane

1.22 grams of oxygen

1.45 grams of oxygen

Answer it was we, the the whole people, who formed the union. and we formed it, not to give the blessings of liberty, but to secure them: not to the half of ourselves and the half of our posterity, but to the whole people- women as well as men.

Use the universal gas law formula: pv = nrt now, the key word in this problem is stp, so at stp, temperature will always be 0 degrees celsius and pressure will always be 1.0 atm or 760 torr. given something is at stp, we already know the two variables pressure and temperature. the other thing we can find with the given information is moles. to find moles, just divide grams by molar mass of the element. 16.0g / 2(16.00)g/mol = 0.5 moles plug the values in to the equation pv = nrt (1.0 atm)( x ) = (0.5 )(0+273=273k) the r value is just a constant for when we use atm, liters, moles and kelvin. i'm assuming you want your answer in liters, but if you dont just move the decimal 3 to the right in your final answer. so when plugging these numbers in to your calculator and trying to find x, you get 11.20665 liters using significant figures the answer is 11.2 liters