Question
What minimum volume of 14.0 M hydrochloric acid would be required to completely dissolve a 600.0 −g iron bar?
I'm getting around 2.3 L but it's saying it's incorrect...
I'm getting around 2.3 L but it's saying it's incorrect...
Answers
Nick
hope this helps
1 OH- plus H+ = HOH or H2O One mole of OH- combines with one mole of H+ to make one mole of water. therefore .50 moles of OH- will combine with .50 moles of H+ to make .50 moles of H2O and the solution will be neutral with a pH of 7.
2. Balance the equation first. 2 HCl + Mg = MgCl2 plus H2(g)
Now get the l0 grams of Mg into moles. l0 grams Mg over 24 grams Mg/mole = .417 moles Mg
Now notice in the balanced equation that 2 moles of HCl reacts with one mole of Mg, therefore we need two times as many moles of HCl as we have of Mg.
2 times .417 moles Mg = .833 moles HCl needed
In any solution Moles solute = Molarity times volume in LITERS
switching this around Volume in Liters = Moles solute over Molarity
Volume HCl needed = .833 moles HCl over 7.0 Molar HCl = .119 Liters or 119 ml
3. Reaction balanced 2 Fe + 6HCl = 2 FeCl3 + 3 H2
Get the 450 grams of iron into moles first. 450 gms Fe over 56 grams Fe per mole = 8.04 moles Fe
Now use the balanced equation as a math platform. Write out the equation on a ;piece of paper.
Over the 2Fe enter 8.04 moles and below the same enter its coefficient of 2 moles
Now over the 6HCl enter X moles and below the same enter its coefficient of 6
Now cross multiply through the equation and solve for X moles of HCl I got 24.1 moles of HCl needed.
Moles HCl = Molarity times Volume in Liters
Volume in liters = Moles HCl over Molarity which = 24.1 moles HCl over 14.0 Molar HCl acid
which = 1.72 Liters HCl acid or 1720 ml. If if the pen held 5 ml of acid, it would have to be refilled about 344 times. No way this would fit into the pen.
1 OH- plus H+ = HOH or H2O One mole of OH- combines with one mole of H+ to make one mole of water. therefore .50 moles of OH- will combine with .50 moles of H+ to make .50 moles of H2O and the solution will be neutral with a pH of 7.
2. Balance the equation first. 2 HCl + Mg = MgCl2 plus H2(g)
Now get the l0 grams of Mg into moles. l0 grams Mg over 24 grams Mg/mole = .417 moles Mg
Now notice in the balanced equation that 2 moles of HCl reacts with one mole of Mg, therefore we need two times as many moles of HCl as we have of Mg.
2 times .417 moles Mg = .833 moles HCl needed
In any solution Moles solute = Molarity times volume in LITERS
switching this around Volume in Liters = Moles solute over Molarity
Volume HCl needed = .833 moles HCl over 7.0 Molar HCl = .119 Liters or 119 ml
3. Reaction balanced 2 Fe + 6HCl = 2 FeCl3 + 3 H2
Get the 450 grams of iron into moles first. 450 gms Fe over 56 grams Fe per mole = 8.04 moles Fe
Now use the balanced equation as a math platform. Write out the equation on a ;piece of paper.
Over the 2Fe enter 8.04 moles and below the same enter its coefficient of 2 moles
Now over the 6HCl enter X moles and below the same enter its coefficient of 6
Now cross multiply through the equation and solve for X moles of HCl I got 24.1 moles of HCl needed.
Moles HCl = Molarity times Volume in Liters
Volume in liters = Moles HCl over Molarity which = 24.1 moles HCl over 14.0 Molar HCl acid
which = 1.72 Liters HCl acid or 1720 ml. If if the pen held 5 ml of acid, it would have to be refilled about 344 times. No way this would fit into the pen.
DrBob222
The work done by Nick is interesting but it doesn't answer the question.
Fe + 2HCl ==> FeCl2 + H2
mols Fe = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Fe to mols HCl. That's mols HCl = twice mols Fe.
Now convert mols HCl to volume.
M HCl = mols HCl/L HCl.
You know M HCl and mols HCl, solve for L HCl. Convert to mL if desired. 2.3 L is way too large by a factor like 2000 time too large.
Fe + 2HCl ==> FeCl2 + H2
mols Fe = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Fe to mols HCl. That's mols HCl = twice mols Fe.
Now convert mols HCl to volume.
M HCl = mols HCl/L HCl.
You know M HCl and mols HCl, solve for L HCl. Convert to mL if desired. 2.3 L is way too large by a factor like 2000 time too large.