Sodium fluoride (naf), sodium fluorosilicate (na2sif6), and fluorosilicic acid (h2sif6) are commercially available in 100.0-lb containers. calculate the mass, in kg, of fluoride ion (f−) in a 100.0-lb container for each of these three compounds. express the masses in kilograms to two significant figures separated by commas.

First let us identify the molar masses:

NaF = 42 g/mol = 42 lb/lbmol

Na2SiF6 = 188 g/mol = 188 lb/lbmol

H2SiF6 = 144 g/mol = 144 lb/lbmol

F = 19 g/mol

There is 1 F in every NaF, 6 F in every Na2SiF, and also 6 F in every H2SiF6. Therefore:

F in NaF = (100 / 42) * 1 * 19 = 45.23 lb = 21 kg

F in Na2SiF6 = (100 / 188) * 6 * 19 = 60.64 lb = 28 kg

F in H2SiF6 = (100 / 144) * 6 * 19 = 79.17 lb = 36 kg

i believe the answer is a.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2