Two problems in stoichiometry
Limiting Reactant and Yield %. See attached file for full problem description.
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<br>PART A.
<br>* In the original reaction the mol ratio of S8 to Cl2 is 1:4.
<br>* The molar weight of sulfur (S) is 32 gr/mol.
<br>*Since we the equation calls for S8, we can see ...
Solution Summary
The solution shows how to find the limiting reactants in solutions.
12 Problems: Stoichiometry, Yield and Substance Remaining
Calculate how many moles of each product would be complete conversion of 1.25 mol of the reactant indicated in boldface, what is the mole ratio used.
C2H3OH (I) + 302(g)  2CO2(g) + 3H2O(g)
N2(g) + O2(g)  2NO(g)
2NaClO2(s) + Cl2(g)  2ClO2(g) + 2NaCl(s)
3H2(g) + N2(g)  2NH3(g)
Calculate the mass of each product that could be produced by complete reaction of 1.55 g of the reactant indicate in the boldface again.
CS2(I) + O2(g)  CO2(g) + SO2(g)
NaNO3(s)  NaNO2(s) + O2(g)
H2(g) + MnO2(s)  MnO(s) +H2O(g)
Br2(l) +Cl2(g)  BrCl(g)
Suppose 5.00g of each reactant is taken. How would you determine which reactant is limiting and what mass of the excess reagent will remain after the limiting reactant is consumed.
Na2B4O7(s) + H2SO(aq) + H2O(l)  H3BO3(s) + Na2SO4(aq)
CaC2(s) + H2O(l)  Ca(OH)2(s) + C2H2(g)
NaCl(s) +H2SO4(l)  HCl(g) + Na2So4(s)
SiO2(s) + C(s)  Si(l) + CO(g)