a 5 g packet of effervescing stomach antacid contains 2.68 g of calcium hydroxide to neutralize stomach acid. What volume of stomach acid (containing 0.010 mol/L HCL) could be neutralized?
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First start with the balanced equation:
Ca(OH)2 + 2 HCl -> CaCl2 + 2 H2O
We only care about the active ingredient in the 5 g packet, which is the 2.68 grams of calcium hydroxide.
Find moles of calcium hydroxide.
2.68 g / (74.09 g/mol) = 0.03617 moles of Ca(OH)2
According to equation, you need 2 moles of HCl for every mole of Ca(OH)2, so you need to calculate how many moles of HCl you can neutralize based on the equation.
0.03617 moles of Ca(OH)2 x (2 moles of HCl/ 1 mole of Ca(OH)2) = 0.0723 moles of HCl will react completely with 0.03617 moles of Ca(OH)2
Next figure out the volume of HCl which contains 0.0723 moles of 0.010 M HCL
(M stands for molarity which is the same as moles/L)
Molarity = moles/ L, therefore L = moles / molarity
L HCl = 0.0723 moles / 0.010 M = 7.23 L
So, 7.23 L of of 0.010 M stomach acid can be neutralized by 2.58 grams of calcium hydroxide.
Ca(OH)2 + 2 HCl -> CaCl2 + 2 H2O
We only care about the active ingredient in the 5 g packet, which is the 2.68 grams of calcium hydroxide.
Find moles of calcium hydroxide.
2.68 g / (74.09 g/mol) = 0.03617 moles of Ca(OH)2
According to equation, you need 2 moles of HCl for every mole of Ca(OH)2, so you need to calculate how many moles of HCl you can neutralize based on the equation.
0.03617 moles of Ca(OH)2 x (2 moles of HCl/ 1 mole of Ca(OH)2) = 0.0723 moles of HCl will react completely with 0.03617 moles of Ca(OH)2
Next figure out the volume of HCl which contains 0.0723 moles of 0.010 M HCL
(M stands for molarity which is the same as moles/L)
Molarity = moles/ L, therefore L = moles / molarity
L HCl = 0.0723 moles / 0.010 M = 7.23 L
So, 7.23 L of of 0.010 M stomach acid can be neutralized by 2.58 grams of calcium hydroxide.