What volume (in L) of a 0.98 M solution of hydrochloric acid, a solution of HCl, can be prepared from the HCl produced by the reaction of 19.03 g of NaCl with an excess of sulfuric acid?
NaCl(s) + H2SO4(aq) = HCl(aq) + NaHSO4(s)
I don't understand how to solve this. Can someone please help explain? Thank you!
2 answers
Never mind... I figured it out.
You have two problems here. The first is to determine the grams of HCl produced from the NaCl. The second problem is to determine the volume of that HCl produced to make a 0.98 M solution of HCl.
First.
.................NaCl(s) + H2SO4(aq) = HCl(aq) + NaHSO4(s)
mols NaCl = g/molar mass = 19.03/58.45 = 0.3256. That will make 0.3256 mols HCl.
Part 2. You want to make 0.98 mols/L HCl.
M = mols/L
mols of 0.98 = mols 0.3256 x L solution. Solve for L = about 3.0 L of the 0.3256 will be 0.98 M
First.
.................NaCl(s) + H2SO4(aq) = HCl(aq) + NaHSO4(s)
mols NaCl = g/molar mass = 19.03/58.45 = 0.3256. That will make 0.3256 mols HCl.
Part 2. You want to make 0.98 mols/L HCl.
M = mols/L
mols of 0.98 = mols 0.3256 x L solution. Solve for L = about 3.0 L of the 0.3256 will be 0.98 M
2 answers