Calculate moles MgCl2 to start = M x L = ?
Calculate moles NaOH to start = M x L = ?
This is a limiting reagent problem; I know that because amounts for BOTH reactants are given. I solve these by solving two simple stoichiometry problems.
First moles MgCl2, ignoring NaOH, second with moles NaOH, ignoring MgCl2.
Convert moles MgCl2 to moles of the product using the coefficients in the balanced equation. Do the same for NaOH to moles of the product. Both answers can't be right; the correct one is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Then convert moles of the product (using the smaller value) to grams. g = moles x molar mass.
Here is a worked example of a stoichiometry problem. This will solve all of your stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
MgCl2(aq) + 2 NaOH(aq) --> Mg(OH)2(s) + 2 NaCl(aq)
How many grams of Mg(OH)2 would form if 50.00 mL of 0.9750M NaOH were combined with 50.00 mL of 0.8820 M MgCl2?
1 answer
1 answer