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stoichiometry example
(how many grams of N2 are necessary to produce 7.5g of NH3) N2 + 3H2 → 2NH3
2NH3 = 7.5g NH3/14g + 3.03g = 0.441 mol
0.411 mol NH3 × 1 mol N2/2 mol NH3 = 0.221 mol N2
0221 mol * (2 × 14)g/1 mol = 0.618g of N2
empirical formula steps
convert to mol
divide by smallest # of mol
round answer to nearest whole #
stoichiometry steps
convert to moles,
multiply # of moles in previous answer by mole of specified element over # of compound in given equation
take that answer and convert it to grams (make sure to multiply by how many are present in equation)
expected % mass loss steps
take the weight of initial compound (multiply by # of compound present)
take weight of mass loss (CO2 and/or H2O)
divide weight of mass loss by initial weight and multiply by 100
calculate % of mass of given element in compound steps
find mass of specified element(s)
divide that mass by the mass of the whole compound
multiply your answer by 100
standard deviation steps
take sum of all the long decimals squared
divide that by # of trials minus 1
square root the whole fraction
limiting reagent steps
convert given elements to mols
multiply the answers by # of reactant present over # of specified compound in equation
whichever is smaller is your answer
calculation of compound in initial mixture
find mass of specified compound within equation
find weight of “mass loss” (CO2 or H2O)
find mass loss with given weights (initial and weight after heating)
multiply the actual mass loss by the mass of specified compound over the weight of “mass loss”
to find the remaining compound:
subtract initial mass by the mass of the previous answer (compound)