Suppose 17. g of hydrochloric cid is mixed with 6.99 g of sodium hydroxide calculate the minimum mass of hydochloric acid taht could be left over by the vhemical reaction

10.62g

Explanation:

First let us generate a balanced equation for the reaction.

HCl + NaOH —> NaCl + H2O

Molar Mass of HCl= 1 + 35.5 = 36.5g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

From the question,

Mass of HCl = 17g

Mass of NaOH = 6.99g

Converting these Masses to mole, we obtain:

n = Mass / Molar Mass

n of HCl = 17/36.5 = 0.4658mol

n of NaOH = 6.99/40 = 0.1748mol

From the question,

1 mole of NaOH requires 1mole of HCl.

Therefore, 0.1748mol of NaOH will also require 0.1748mol of HCl.

But we were told that 17g( i.e 0.4658mol) of HCl were mixed.

Therefore, the unreacted amount of HCl = 0.4658 — 0.1748 = 0.291mol

Converting this to mass, we have:

Mass of HCl = n x molar Mass

Mass of HCl = 0.291 x 36.5

Mass of HCl = 10.62g

Therefore the left over Mass of HCl is 10.62g