74.4 ml of millimeters needed to react with 15 grams of baking soda
<u>Explanation:</u>
Given data
n C8 H8 07= 0.8 M
m NaHCO3=15 g
<u>Solving the problem:</u>
From the equation we get,
3x C6 H8 07 = NaHC03
Finding the molar mass for the baking soda
NaHCO3= 23+1+12+3 x16= 84 g / mol
so,
n NaHCO3= m NaHCO3/ M NaHCO3
And then,
n C6 H8 07 =1/3 × n NaHCO3
From the definition of the molarity we get
V C6 H8 O 7=n C6 H8 07/C6 H8 07
=n NaHCO3/3 × C6 H8 07 =
m NaHCO3/ M NaHCO3/ 3 × C6 H8 07
Calculating the equation we get,
V C6 H8 O 7= 15/84/3 ×0.8=5/84 ×0.8= 0.0744L= 74.4 ml
74.4 ml of millimeters needed to react with 15 grams of baking soda
n C8 H8 07= 0.8 M
m NaHCO3=15 g
Solving the problem:
From the equation we get,
3x C6 H8 07 = NaHC03
Finding the molar mass for the baking soda
Nahco3= 23+1+12+3 x16= 84 g / mol
so,
n NaHCO3= m NaHCO3/ M NaHCO3
And then,
n C6 H8 07 =1/3 × n NaHCO3
From the definition of the molarity we get
V C6 H8 O 7=n C6 H8 07/C6 H8 07
=n NaHCO3/3 × C6 H8 07 =
m NaHCO3/ M NaHCO3/ 3 × C6 H8 07
Calculating the equation we get,
V C6 H8 O 7= 15/84/3 ×0.8=5/84 ×0.8= 0.0744L= 74.4 ml
74.4 ml of millimeters needed to react with 15 grams of baking soda