Respuesta :
Explanation:
We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.
[tex]H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)[/tex]
Next we will calculate how many moles of [tex]H_2SO_4[/tex] are present in 85.00 mL of 1.500 M sulfuric acid.
As, Molarity = [tex]\frac{\text{moles of solute}}{\text{liters of solution }}[/tex]
1.500 M = [tex]\frac{n}{0.08500 L }[/tex]
n = 0.1275 mol [tex]H_2SO_4[/tex]
Now set up and solve a stoichiometric conversion from moles of [tex]H_2SO_4[/tex] to grams of [tex]NaHCO_3[/tex]. As, the molar mass of [tex]NaHCO_3[/tex] is 84.01 g/mol.
[tex]0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})[/tex]
= 21.42 g [tex]NaHCO_3[/tex]
So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.
15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid.
Balanced chemical equation:
[tex]H_2SO_4(aq)+2NaHCO_3(s)---- > Na_2SO_4(aq)+2H_2O(l)+2CO_2(g)[/tex]
Calculation for number of moles:
Molarity = Number of moles of solute / Volume of solution
1.500 M = n / 0.08500 L
n = 0.1275 moles of H2SO4
According to mole-ratio concept:
Molar mass of sodium bicarbonate = 84.01 g/mol
[tex]\text{ Mass of } NaHCO_3=0.1275 mol *\frac{2 mol}{1 mol} *\frac{84.01g}{1 mol} \\\\\text{ Mass of } NaHCO_3}=21.42 \text{ g of } NaHCO_3[/tex]
So, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.
Find more information about Number of moles here:
brainly.com/question/13314627
