Question

37.3 mL of a 0.129 M Na2CO3 solution completely react with a 0.150 M HNO3 solution according to the following balanced chemical equation: Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l) What mass (in grams) of carbon dioxide is formed?

Answer 1

Balanced chemical equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

lets calculate the mol of Na2CO3

volume , V = 37.3 mL

= 3.73*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.129*3.73*10^-2

= 4.812*10^-3 mol

According to balanced equation

mol of CO2 FORMED = (1/1)* moles of Na2CO3

= (1/1)*4.812*10^-3

= 4.812*10^-3 mol

This is number of moles of CO2

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

use:

mass of CO2,

m = number of mol * molar mass

= 4.812*10^-3 mol * 44.01 g/mol

= 0.2118 g

Answer: 0.212 g