Question

Na2SiO3 + 8 HF -> H2SiF6 + 2NaF + 2H20

a. how many moles is 220 grams of Na2SiO3?

b. How many grams of NaF form when .560 mol of HF react with 220 g of Na2SiO3? FInd LR.

c. What is the excess reactant and how many moles of excess reactant remain?

Please show and explain every step. Thank you.

Answer 1

The balanced equation is Na2SiO3 + 8 HF -> H2SiF6 + 2NaF + 2H20

Molar mass(g/mole)            122             20        144          42        18

(a) Number of moles of Na2SiO3 , n = mass/molar mass

                                                    = 220g / 122(g/mole)

                                                    = 1.80 mole

(c) From the balanced reaction ,

8 moles of HF reacts with 1 mole of Na2SiO3

0.560 moles of HF reacts with (0.560 x 1)/8 = 0.07 moles of Na2SiO3

So 1.80 - 0.07 = 1.73 moles of Na2SiO3 left unreacted and is the excess reactant

Since all the mass of HF reacted it is the limiting reactant.

(b) Again from the balanced reaction

8 moles of HF produces 2 moles of NaF

0.560 moles of HF produces (0.560 x 2 ) / 8 =0.14 moles of NaF

So mass of NaF produced , m = number of moles x molar mass

                                            = 0.14 moles x 42 g/mole

                                            = 5.88 g