Asked by Reeseamelia
How many milliliters of a 4.5 M HCl solution would be needed to react completely with 5.8 g of aluminum?
I know the answer is somewhere around 143 mL HCl, but I CANNOT figure out how to find that answer. I have done a conversion from g of Al to mL of HCl and I cannot find the real answer.
I know the answer is somewhere around 143 mL HCl, but I CANNOT figure out how to find that answer. I have done a conversion from g of Al to mL of HCl and I cannot find the real answer.
Answers
Answered by
DrBob222
1. Write and balance the equation.
2Al + 6HCl ==> 2AlCl3 + 3H2
2. Convert 5.8 g Al to mols. mol = grams/atomic mass = 5.8/27 = approx 0.2 but you need a more accurate answer than that estimate.
3. Using the coefficients in the balanced equation, convert mols Al to mols HCl. That's 0.2 mol Al x (6 mols HCl/2 mols Al) = 0.2 x 6/2 = approx 0.6. Recalculate this one also.
4. Finally, M HCl = mols HCl/L HCl. You know M HCl and mols HCl, solve for L HCl and convert to mL.
2Al + 6HCl ==> 2AlCl3 + 3H2
2. Convert 5.8 g Al to mols. mol = grams/atomic mass = 5.8/27 = approx 0.2 but you need a more accurate answer than that estimate.
3. Using the coefficients in the balanced equation, convert mols Al to mols HCl. That's 0.2 mol Al x (6 mols HCl/2 mols Al) = 0.2 x 6/2 = approx 0.6. Recalculate this one also.
4. Finally, M HCl = mols HCl/L HCl. You know M HCl and mols HCl, solve for L HCl and convert to mL.
Answered by
Anonymous
no
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