If a total of 13.5 mol of NaHCO3 and 4.5 mol of C6H8O7 react, how many moles of CO2 and Na3C6H5O7 will be produced?
3NaHCO3(aq) + C6H8O7(aq) ƒ{ƒ®3CO2(g) + 3H2O(s) +Na3C6H5O7(aq)
i really hate chemistry it hurts my brain i have problems with all the techinical thingys so im guessing again yaay me
is the answer 151.9
No.
The coefficients in the equation will tell you. Have you written the balanced equation?
To find out how many moles of CO2 and Na3C6H5O7 will be produced, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is:
3NaHCO3(aq) + C6H8O7(aq) → 3CO2(g) + 3H2O(s) + Na3C6H5O7(aq)
From the balanced equation, we can determine the stoichiometric ratio between the reactants and products. In this case, the ratio between NaHCO3 and CO2 is 3:3, and the ratio between C6H8O7 and Na3C6H5O7 is 1:1.
Given that there are 13.5 mol of NaHCO3 and 4.5 mol of C6H8O7, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed first, thereby determining the maximum amount of product that can be formed.
To find the limiting reactant, we compare the moles of the reactants to their stoichiometric coefficients. Divide the number of moles of each reactant by its stoichiometric coefficient to find the number of moles per coefficient:
NaHCO3: 13.5 mol / 3 = 4.5 mol per coefficient
C6H8O7: 4.5 mol / 1 = 4.5 mol per coefficient
In this case, both reactants have the same number of moles per coefficient. Therefore, neither NaHCO3 nor C6H8O7 is limiting; both are in excess.
Now we can calculate the moles of CO2 and Na3C6H5O7 produced using the stoichiometry of the balanced equation.
Since the stoichiometric ratio between NaHCO3 and CO2 is 3:3, for every 3 moles of NaHCO3, 3 moles of CO2 are produced. Thus, with 4.5 mol of NaHCO3 in excess, we will have 4.5 mol of CO2 produced.
Similarly, since the stoichiometric ratio between C6H8O7 and Na3C6H5O7 is 1:1, for every 1 mole of C6H8O7, 1 mole of Na3C6H5O7 is produced. Thus, with 4.5 mol of C6H8O7 in excess, we will have 4.5 mol of Na3C6H5O7 produced.
Therefore, when 13.5 mol of NaHCO3 and 4.5 mol of C6H8O7 react, 4.5 mol of CO2 and 4.5 mol of Na3C6H5O7 will be produced.