I've been at this short chem problem for hours. Any help please?!?

Question

Calculate the molarity of a solution of HI if 86.82 mL of the HI solution neutralize 0.741 g of Na2CO3.

Answer 1

you have the reaction

2 HI + NA2CO3 ---> CO3H2 + 2NaI

NA2CO3 moecular weight = 2*23+12+3*16= 106 g

so 0.741 correspond to a molarity of 0.741/106 = 0.007 moles

as you have 2 NaI for 1 NA2CO3 the mole of Na I is 0.007*2 =0.014 moles

so you have 0.014 mole in 86.82ml = 0.08682 liter

and in a liter you have 0.014/0.08682 =0.16M

Answer 2

Here's how you do the problem. I'll leave the math to you.

Write the chemical equation for the neutralization of sodium carbonate by hydrogen iodide and balance it. Calculate the number of moles of sodium carbonate neutralized. From this calculate the mass of hydrogen iodide needed to do the job. How many moles is this? Divide this by .08682 to get the molarity.

Answer 3

.741 g Na2CO3 has 0.0069906 moles CO3-2

each mole of CO3-2 will need 2 moles of H+

so you will need .0139 moles

if you used 86.86 ml HI

that yielded 0.0139 moles H+ = 86.86ml (Xmole/1000ml)

Solving for X gives 0.16 M