Chemistry question, equation about a experiment how many miles

Question

Hi I need help with question 46 and 47 thank you

Sodium thiosulphate (Na2S2O3) (a solid) was reacted with 500cm3 of hydrochloric acid in solution to form a precipitate (sulphur). The rate of this reaction can be studied by measuring the time that it takes for a small fixed amount of precipitate to form under different conditions.

The equation for this reaction is shown below.

Na2S2O3 + HCl→NaCl + S + SO2 + H2O

45) Balance the equation by writing in the box below (M4)

46) At the end of the reaction the amount of sulphur was weighed at 64000mg. What is the mass of sodium thiosulphate that reacted? (M4 and D3)

(NB. Relative atomic masses: Sodium – 23; Sulphur – 32; Oxygen – 16; Hydrogen – 1; Chlorine – 35)

47) What is the concentration of the hydrochloric acid solution? (D3)

Answer 1

In order to answer this, one must first assume that the HCl is present in excess, and that the reaction goes to completion.

Na₂S₂O₃ + 2HCl ===> 2NaCl + S + SO₂ + H₂O ... balanced equation

moles S formed = 64000 mg x 1 g/1000 mg x 1 mole/32 g = 2 moles S formed

moles Na₂S₂O₃ originally present = 2 moles S x 1 moles Na₂S₂O₃/mole S = 2 moles Na₂S₂O₃

molar mass Na₂S₂O₃ = 23+23+32+32+ 16+16+16 = 158 g/mole

mass Na₂S₂O₃ = 2 moles x 158 g/mole = 316 g of Na2S2O3

Concentration of HCl:

From balanced equation, 2 moles HCl reacts with 1 mole Na₂S₂O₃

moles HCl = 2 moles Na₂S₂O₃ x 2 moles HCl/mole Na₂S₂O₃ = 4 moles HCl used

molarity of HCl = moles/liter = 4 moles/0.5 L = 8 moles/liter = 8 M