What volume of 1.0 M HCl would react completely with 0.20 grams of NaC2H3O2*3H2O?
It sounds like an equation I would set up as this and would use mole ratios to calculate the answer:
HCl + NaC2H3O2*3H2O -->
But I am confused as to what I would put as the products, let alone how to even begin the problem. Any help is greatly appreciated :)
To start solving this problem, you need to understand the balanced chemical equation for the reaction between HCl (hydrochloric acid) and NaC2H3O2*3H2O (sodium acetate trihydrate). The equation is:
HCl + NaC2H3O2*3H2O -> NaCl + C2H4O2 (acetic acid) + 3H2O
In this reaction, HCl reacts with NaC2H3O2*3H2O to form NaCl, acetic acid (C2H4O2), and water (H2O). Note that the sodium acetate trihydrate decomposes to form acetic acid and water.
Now, let's proceed with the calculation. First, you need to convert the given mass of NaC2H3O2*3H2O (0.20 grams) to moles. To do this, you'll need to know the molar mass of NaC2H3O2*3H2O:
Molar mass of Na = 22.99 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol
So, the molar mass of NaC2H3O2*3H2O = (22.99 g/mol * 1) + (12.01 g/mol * 2) + (1.01 g/mol * 5) + (16.00 g/mol * 5) + (3 x 1.01 g/mol) + (3 x 16.00 g/mol) = 136.09 g/mol
Now, calculate the moles of NaC2H3O2*3H2O using the given mass:
Moles = Mass / Molar mass = 0.20 g / 136.09 g/mol ≈ 0.00147 mol
Next, since the balanced equation shows a 1:1 mole ratio between HCl and NaC2H3O2*3H2O, the moles of HCl needed to react completely with the given moles of NaC2H3O2*3H2O will also be 0.00147 mol.
Finally, to find the volume of 1.0 M (1.0 mol/L) HCl needed, you can use the equation:
Volume (L) = Moles / Concentration (M)
Substituting the given values:
Volume = 0.00147 mol / 1.0 M ≈ 0.00147 L (or 1.47 mL)
Therefore, approximately 0.00147 L (or 1.47 mL) of 1.0 M HCl would react completely with 0.20 grams of NaC2H3O2*3H2O.