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)
12 Problems regarding Stoichiometry, Theoretical Yield and Substance Remaining are solved in fine detail.