What mass in grams of fructose is contained in 325 ml of a 1.5 m fructose solution
this is a limiting reagent question.
1) determine which substance is the limiting reagent, by converting both masses into moles (using applicable molecular masses, derived from the periodic table), and comparing the relative quantities we actually have, to the relative quantities the balanced chemical equation says we need.
2) use the balanced chemical equation to find the applicable mole ratio, and convert from moles of the limiting reagent into moles of product.
3) convert moles of product into mass of product, again using molecular masses derived from the periodic table.
(10g c3h8)(1mol c3h8 / 44.11g c3h8)
= 0.23mol c3h8 available
(10g o2)(1mol o2 / 32.00g o2)
= 0.31mol o2 available
the balanced chemical equation tells us that we need 5 moles of o2 for every 1mol of c3h8
(0.23mol c3h8)(5mol o2 / 1mol c3h8)
= 1.2mol o2 required
comparing this to the amount of o2 we actually have available, it should be apparent that we don't have enough; we're going to run out of o2 first, so that's our limiting reagent.
the balanced chemical equation tells us that 3 moles of co2 will be produced for every 5 moles of o2 reacted.
(1.2mol o2)(3mol co2 / 5mol o2)
= 0.72mol co2
(0.72mol co2)(44.01g co2 / 1mol co2)
= 32g co2
key idea to remember: the quantities of reactants and products are related through the mole ratios given by the balanced chemical equation of the reaction
if it was propogating through completely free space, it would be completely lossless and would therefore continue without change for infinity.