Asked by Amanda
How much 0.5 M HCl is needed to neutralize 5.5 kg of ammonia dissolved in 6 gal of water?
Answers
Answered by
DrBob222
NH3 + HCl ==> NH4Cl
5.5 kg = 5500 g
Convert to moles.
moles = grams/molar mass = ?
1 mole NH3 requires 1 mol HCl; therefore, moles NH3 = moles HCl.
M HCl = moles HCl/L HCl.
You know M HCl and moles HCl, solve for L HCl
5.5 kg = 5500 g
Convert to moles.
moles = grams/molar mass = ?
1 mole NH3 requires 1 mol HCl; therefore, moles NH3 = moles HCl.
M HCl = moles HCl/L HCl.
You know M HCl and moles HCl, solve for L HCl
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