Question

H3c6h5o7(aq) + 3nahco3(aq) → 3co2(g) + 3h2o(l) + na3c6h5o7(aq) calculate the number of grams of baking soda (nahco3; molar mass 84.00661 g/mol) that will react with 30.0 ml of 1 m citric acid.

Answer 1

Determine the number of moles of citric acid.

n of citric acid = (30 mL)(1 L / 1000 mL)(1 mol/L)
= 0.03 mol of citric acid

It is seen in the equation that each mole of citric acid reacts with 3 moles of baking soda.

n of baking soda = (0.03 mol of citric acid)(3 mols baking soda/1 mol of citric acid)

n of baking soda = 0.09 mol of baking soda

mass of baking soda = (0.09 mol of baking soda)(84.00661 g/mol)
mass of baking soda = 7.56 g of baking soda

Answer: 7.56 g