Question

Consider two reactants (A and B) that when mixed form a precipitate C, which can be filtered, dried, and then weighed. Carlos attempts to find the stoichiometric ratios in which A and B mix, by mixing different volumes of the same concentration of solutions of A and B (0.50 M). He records the following data:
Run Vol A (mL) Vol B (mL) Mass C (g)
1 10.0 10.0 1.17
2 20.0 10.0 2.33
3 30.0 10.0 3.50
4 40.0 10.0 3.50

Assuming that the products of the reaction are C and D, and that the stoichiometric coefficients for C and D are 1 and 2 respectively, the balanced chemical reaction will be

( ) A + ( ) B â 1 C + 2 D

Answer 1

Answer:

3A     +     B      →     C      +      2D

Explanation:

Recall that:

$\text{molarity} = \dfrac{no \ of \ moles}{volume \ of \ the \ solution}$

So;

$\text{no \ of \ moles= molarity * volume \ of \ the \ solution}$

From the above table given in the question; we can have the following table:

Run            moles of A            moles of B     Limiting reagent

1                 0.5×  10⁻²                 0.5×  10⁻²           A and B are equal

2               1.0 0.5×  10⁻²             0.5×  10⁻²          A

3.               1.5×  10⁻²                   0.5×  10⁻²          A

4               2.0×  10⁻²                    0.5×  10⁻²         B

Also, provided that the stoichiometric coefficients for C and D = 1 & 2 respectively.

Then, the stiochiometric ratio for A:B = $\dfrac{1.5\times 10^{-2}}{0.5 \times 10^{-2}}$

$= \dfrac{3}{1}$

Thus, the balanced equation is:

3A     +     B      →     C      +      2D