Exam 3 Flashcards
understanding the molecular and mass information in a balanced equation
- balanced chm equation gives amnt of reactant and product molecules that participate in a chemical run
- coefficients in a balanced eq. give the relative # of molecules
- all atoms present in the reactants are accounted for in the products
- 1 molecule of ethanol reacts w/ 3 molecules of O2 to produce 2 molecules of CO2 and 3 molecules of H2O
- 1 mole of ethanol reacts w 3 moles O2 to produce 2 moles of CO2 and 3 moles of H2O
Cu2S(s) + Cu2O(s) →Cu(s) + SO2(g) (unbalanced)
After balancing the reaction, determine how many dozens of copper atomscan be produced from a dozen molecules of the sulfide and two dozen molecules of the oxide of copper(I). Also, how many molesof coppercan be produced from one mole of the sulfide and two moles of the oxide?
6 dozens and 6 moles
relationship between moles of reactants and moles of products
- balanced eq can predict moles of products that a given # of moles of reactants will yield
- 2 moles of H2O yield 2 moles of H2 and 1 mole of O2
- 4 moles of H2O yield 4 moles of H2 and 2 moles of O2
mole ratio
allows us to convert from moles of 1 substance in a balanced eq to moles of a second substance in the eq
coefficient of unknown / coefficient of known
Consider the following balanced equation:Na2SiF6(s) + 4Na(s) → Si(s) + 6NaF(s)
How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na2SiF6?
5.25 moles NaF
Propane, C3H8, is a common fuel used for heating in rural areas. Predict the number of moles of CO2 formed when 3.74 moles of propane is burned in excess oxygen.
oxygen.
C3H8+ 5O2→ 3CO2+ 4H2O
11.2 moles
steps for calculating the masses of reactants and products in chm rxns
- balance the eq for the rxn
- convert the masses of reactants/products to moles
- use balanced eq to set up appropriate mole
- use mole ratio(s) to calculate the # of moles of desired reactant or product
- convert moles back to mass
For the following unbalanced equation: Cr(s) + O2(g) → Cr2O3(s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen?
21.9 g Cr2O3
Tin(II) fluoride is added to some dental products to help prevent cavities. Tin(II) fluoride is prepared according to the following equation:
Sn(s) + 2HF(aq) SnF2(aq) + H2(g)
How many grams of tin(II) fluoride can be produced from 55.0 g of hydrogen fluoride if there is plenty of tin available to react?
215 g
Consider the following reaction:
PCl3+ 3H2O H3PO3+ 3HCl
What mass of H2O is needed to completely react with 20.0 g of PCl3?
7.87g H2O
Consider the following reaction where X represents an unknown element:
6X(s) + 2B2O3(s) B4X3(s) + XO2(g)
If 175 g of X reacts with diborontrioxide to produce 2.43 moles of B4X3, what element does X represent?
C
limiting reactant mixture
N2 (g) + 3 H2 (g) -> 2 NH3 (g)
- limiting reagent is the reactant that runs out 1st and limits the amnt of products formed
- H2 is limiting reactant for rxn
limiting reactants
- determine which reactant is limiting to calculate the amnt of products tht will be formed
- methane and water will react to form products according to the eq: CH4 + H2O -> 3 H2 + CO
a) H2O molecules are used up first, leaving 2 unreacted CH4 molecules - amnt of products that can form is limited by the water
- water is limiting reactant
- methane is in excess
steps for solving stoichiometry problems involving limiting reactants by solving for limiting reactant first (Approach A)
- write + balance eq for rxn
- convert known masses of reactants to moles
- using the #s of moles of reactants + the appropriate mole ratios, determine which reactant is limiting
- using the amnt of limiting reactant and appropriate mole ratios, compute the # of moles of desired product
- if required: convert from moles of product to grams of product, using the molar mass
example 3 Approach A:
You react 10.0 g of A with 10.0 g of B. Given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol, what will be the mass of the product produced?
A + 3B -> 2C
8.33 g C
when solving for the amount of excess left over we must find out how much was __ - how much we __ __
needed; started with
steps for solving stoichiometry problems involving limiting reactants by determining the # of moles of products first (Approach B)
- write + balance eq for rxn
- convert known masses of reactants to moles
- using approp mole ratios, compute # of moles of product formed if each reactant was consumed
a) choose least # of moles of product formed - if required: convert moles of product to grams of product, using the molar mass
Consider the equation: A + 3B 4C. If 3.0 moles of A is reacted with 6.0 moles of B, which of the following is true after the reaction is complete?
A is the leftover reactant because you only need 2 moles of A and have 3 moles
How many grams of NH3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process?
N2+ 3H2→ 2NH3
3.65 g NH3
percent yield
- important indicator of the efficiency of a particular lab/industrial rxn
- comparison of the actual yield of a product to its theoretical yield
% yield = actual yield / theoretical yield x 100%
theoretical yield
- the max amnt of a given product that can be formed when the limiting reactant is completely obtained
- the amnt of a product actually obtained
actual yield
the amnt actually produced, of a rxn, is usually less than the max. expected (theoretical yield)
You react 10.0 g of A with 10.0 g of B. Given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol, what will be the mass of the product produced?
A + 3B 2C
- We determined that 8.33 g C should be produced. 7.23 g of C is what was actually made in the lab. What is the percent yield of the reaction?
(7.23g C / 8.33g C) x 100% = 86.8%
find the percent yield of a product if 1.50g of SO3 is made from 1.00g of O2 + excess sulfur
2 S + 3 O2 -> 2 SO3
89.8%