B. How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda?

74.4 ml of millimeters needed to react with  15 grams of baking soda

Explanation:

Given data

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

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

i know insulin is primary but i dont know the others im sorry

explanation: