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

Answer: 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