Answer:
[tex]\boxed{\text{1000 g}}[/tex]
Explanation:
We know we will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 84.01
H₂SO4 + 2NaHCO₃ ⟶ Na₂SO₄ + 2CO₂ + 2H₂O
n/mol: 6
1. Use the molar ratio of NaHCO₃ to calculate the moles of NaHCO₃.
[tex]\text{Moles of NaHCO$_{3}$ = 6 mol H$_{2}$SO$_{4}$} \times \dfrac{\text{2 mol NaHCO$_{3}$}}{\text{1 mol H$_{2}$SO$_{4}$}}\\=\text{12 mol NaHCO$_{3}$}[/tex]
2. Use the molar mass of NaHCO₃ to calculate the mass of NaHCO₃.
[tex]\text{Mass of NaHCO$_{3}$ = 12 mol NaHCO$_{3}$} \times \dfrac{\text{84.01 g NaHCO$_{3}$}}{\text{1 mol NaHCO$_{3}$}}\\\\= \text{1000 g NaHCO$_{3}$}[/tex]
You must use [tex]\boxed{\textbf{1000 g}}[/tex] of NaHCO₃.
