Please, Please help

A flask containing 550 mL of 0.75 M H2SO4 was accidentally knocked to the floor.
How many grams of NaHCO3 do you need to put on the spill to neutralize the acid according to the following equation?
H2SO4(aq)+2NaHCO3(aq)→Na2SO4(aq)+2H 2O(l)+2CO2(g)

Express your answer using two significant figures.
m= g

We need 69 grams of NaHCO3

Explanation:

Step 1: Data given

Volume = 550 mL = 0.550 L

Molarity H2SO4 = 0.75 M

Step 2: The balanced equation

H2SO4(aq) + 2NaHCO3(aq) → Na2SO4(aq) + 2H2O(l) + 2CO2(g)

Step 3: Calculate moles of H2SO4

Moles H2SO4 = molarity * volume

Moles H2SO4 = 0.75 M * 0.550 L

Moles H2SO4 =  0.4125 moles H2SO4

Step 4: Calculate moles NaHCO3

For 1 mol H2SO4 we need 2 moles NaHCO3 to produce 1 mol Na2SO4 and 2 moles H2O and 2 Moles CO2

For 0.4125 moles H2SO4 we need 2*0.4125 = 0.825 moles NaHCO3

Step 5: Calculate mass NaHCO3

Mass NaHCO3 = moles * molar mass

Mass NaHCO3 = 0.825 moles * 84.0 g/mol

Mass NaHCO3 = 69.3 grams ≈ 69 grams

We need 69 grams of NaHCO3

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