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ankara55t · Dec 7, 2009, 05:23 AM

In NaHC03 + HCl ----yields NaCl + H20 + C02

if we use 6.0ml of a 6.0M HCl, to calculate the # of moles and # of grams of HCl, do I do the following: 6.0ml X 6moles/1000ml. To get the # of moles then just multiply this by the molecular weight of HCl to get the # of grams?

Perito · Dec 7, 2009, 07:05 AM

Originally Posted by ;

If we use 6.0ml of a 6.0M HCl, to calculate the # of moles and # of grams of HCl, do I do the following: 6.0ml X 6moles/1000ml. to get the # of moles then just multiply this by the molecular weight of HCl to get the # of grams?

This is your equation:

$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$

If you count atoms, you'll find that the equation is balanced. You have the same number of Na, H, C, O, and Cl on both sides of the equation. Therefore you can say the following:

1 mole of NaHCO3 will react with one mole of HCl to give 1 mole of NaCl, 1 mole of water, and 1 mole of carbon dioxide.

The first thing to do is to figure out how much HCl you have.

$6.0\,\cancel{mL} \, \times \, \frac {6.0\,moles}{\cancel{Liter}} \,\times\, \frac {1\,\cancel {Liter}}{1000\,\cancel{mL}}=3.6\times 10^{-2}\, moles$

The molecular weight of HCl is 36.46 g/mole

$3.6\times\10^{-2}\,\cancel {moles} \,\times\, \frac {36.46\,g}{\cancel {mole}} = 1.31\,g$

Note how the units cancel. This is a general way to solve any of these problems. The units can tell you what to do next.

Unknown008 · Dec 7, 2009, 08:27 PM

In other words, it's a 'yes'.