A mixture of 4.20 g of $NaHCO X_{3}$ and 2.65 g of $Na X_{2} CO X_{3}$ is treated with excess dilute $H X_{2} SO X_{4}$ and all the $CO X_{2}$ evolved is passed into excess lime water. Atomic masses: Na 23, C 12, O 16, H 1, Ca 40. Calculate the mass of $CaCO X_{3}$ precipitated in grams, correct to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

Answer: 7.5 g

NaHCO X_{3}

(M = 84): 4.20/84 = 0.0500 mol, and

Na X_{2} CO X_{3}

(M = 106): 2.65/106 = 0.0250 mol. Each mole of either salt gives one mole of

CO X_{2}

, so 0.0750 mol

CO X_{2}

forms.

CO X_{2} + Ca (OH) X_{2} CaCO X_{3} + H X_{2} O

gives 0.0750 mol of

CaCO X_{3}

(M = 100), i.e. 7.50 g.