Answer to Question #106417 in General Chemistry for Whitney

Question

2HF+1NaSiO3---->2NaF+1H2SiO3
how many grams of H2SiO3 would be made from 0.54 mol of HF?
How many moles of Na2SiO3 would be needed to react with 8 mol of HF?
How many moles of H2SiO3 would be made with 101 grams of NaF?
How many moles of NaF would be made from 25 mol of HF?
How many grams of H2SiO3 would be made from 75.0 grams of HF?

Answer 1

0.54 mol of HF makes 0.54/2 = 0.27 mol H2SiO3 or m = 0.27* 78,10 = 21,09 g;

8 mol of HF reacts with 4 mol of Na2SiO3;

101 grams of NaF represents 101/41,99 = 2.41 mol and is produces 2.41/2 = 1,20 mol H2SiO3;

25 mol of HF produce 25 mol of NaF

75.0 grams of HF equal to 75.0/20.01 = 3.75 mol and creates 3.75/2 = 1.87 mol of H2SiO3 or m = 1.87*78.10 = 146.36 g!