Chapter 3

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of 5.0 grams of O2 with 6.0 grams of S. What is the % yield of SO3 in this experiment? S (s) + O2 (g) → SO3 (g) (not balanced)

-balance -find grams of all three -if 5 gram of O2, then 3.3 g of S. since 3.3<6g S then O2 is limiting reactant!! * if 3.3>6, then need to calculate with S instead. -so then product is 8.3 g <--*THEORETICAL YIELD* -7.9 is ACTUAL YIELD -ACTUAL/THEO=% YIELD

1 mol, 2 carbon atoms. 2 mol, 4 carbon atoms C2

1 mol, 2 carbon atoms. 2 mol, 4 carbon atoms C2

finding mass % of an element

100/amu x (element's mass x # of atoms)

How many molecules of CH4 are in 48.2 g of this compound?

48.2/16=A A x 6.02x10^23=answer

What is the empirical formula of a compound that contains 29% Na, 41% S, and 30% O by mass?

assume 100 grams. 29g x (1/23)=Na gram 41g x (1/32.1)=S gram 30g x (1/16)=O gram divide by smallest gram round to nearest whole number double to get rid of fraction

equation for the reaction that occurs when element is burned in air

element + O2 -> CO2 + H2O

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N2(g) + 3H2(g) → 2NH3(g) If the reaction yield is 87.5%, how many moles of N2 are needed to produce 3.00 mol of NH3?.

if 100%, 1.5 N2 is needed for 3 mol. since 87% is yielded, 100/87.5 x 1.5=1.7 mol N2

How many oxygen atoms are contained in 2.74 g of Al2(SO4)3?

total grams x 1/342.17 x 12(oxygen mol) x 6.02x10^23

finding mol of a compound in compound...

sample/actual=A mol x A= molA

ammonium carbonate

(NH4)2CO3

There are __________ atoms of oxygen are in 300 molecules of CH3CO2H.

600

How many moles of sodium carbonate contain 1.773 ×10^17 carbon atoms?

carbon atoms/6.02x10^23=moles of sodium carbonate Na2CO3