  How many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 79.0 mL of 0.750 M AgNO3 solution

Question

How many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 79.0 mL of 0.750 M AgNO3 solution

Step 1

The molarity of a solution is the ratio of number of moles of a substance to the volume of solution in liters. The expression for molarity is labelled in equation (1) in which M is the molarity, n is the number of moles of solute and  V is the volume in liters.

The relation between millilitres (mL) and liters (L) is written in equation(2).

The number of moles of a substance is the ratio of its given mass by molar mass of the substance. It is expressed in equation (3) in which n is the number of moles, m is the given mass and M is the molar mass. help_outlineImage TranscriptioncloseМ-. V(L) (1) .. (2) 1 L 1000 mL (3) n = E Z l fullscreen
Step 2

Convert 79.0 mL in L by using the equation (2) as shown below. help_outlineImage Transcriptionclose1 L 79 mL 79 mL 1000 mL =0.079 L fullscreen
Step 3

The reaction of sodium carbonate and silver nitrate is written below.

2AgNO3(aq) + Na2CO3(aq) → Ag2CO3 (s) +2NaNO3(aq)

The number of moles of sil... help_outlineImage Transcriptionclosen 0.750 M= 0.079 L n 0.0592 mol fullscreen

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