If 200 mL of 4.3500 M aqueous HCl reacts stoichiometrically according to the balanced equation, how many milliliters of 4.54 M aqueous NaCl are produced?
Na2CO3(aq) + 2HCl(aq) → CO2(g) + 2NaCl(aq) + H2O(l)
MM
HCl 36.461
NaCl 58.442
millimoles HCl = mL x M = ?
millimoles NaCl = millimoles HCl
millimoles NaCl = mL x M. You know millimoles NaCl and M, solve for mL. Post your work if you get stuck.
I THINK this answers the question you have tried to ask but let me point out that such answer doesn't mean anything. You already have a solution of HCl pf 200 mL so whavever you calculate for the NaCl doesn't mean anything.
ur a lifesaver drbob222 i appreciate u <3
To find out how many milliliters of 4.54 M aqueous NaCl are produced, you'll need to use the stoichiometry of the balanced equation and the given information.
First, let's determine the number of moles of HCl in 200 mL of 4.3500 M HCl.
moles of HCl = volume (in L) x molarity
moles of HCl = 200 mL x (1 L/1000 mL) x 4.3500 mol/L
Next, using the stoichiometry of the balanced equation, we can see that for every 2 moles of HCl, we get 2 moles of NaCl. Therefore, the moles of NaCl produced will be the same as the moles of HCl.
Now, let's calculate the number of milliliters of 4.54 M aqueous NaCl produced.
volume (in L) of NaCl = moles of NaCl / molarity
volume (in L) of NaCl = moles of HCl x (1 L/1000 mL) x (1/2) x 4.54 mol/L
Finally, convert the volume back to milliliters:
volume (in mL) of NaCl = volume (in L) of NaCl x 1000 mL/L
Plug in the values and calculate the result.