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

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol