The ratio of Mg to F is constant.
f2 / m2 = f1 / m1
To find the second mass of fluorine (f2),
rearrange and substitute appropriate values.
Upon decomposition, one sample of magnesium fluoride produced 1.65kg of magnesium and 2.58kg of fluorine. A second sample produce 1.38kg of magnesium. How much fluorine (in grams) did the second sample produce?
5 answers
F/Mg = 2.58/1.65
So, given 1.28kg Mg, we should have
1.28*(2.58/1.65) = 2.00kg F
So, given 1.28kg Mg, we should have
1.28*(2.58/1.65) = 2.00kg F
You need to create an identity to solve for the mass of fluorine:
1.65 kg of Mg=2.58 kg of F
and
1.38 kg of Mg=x kg of F
So,
x kg of F/1.38kg of Mg=2.58 kg of F/1.65 kg of Mg
Solve for x kg of F,
x kg of F=(2.58 kg of F/1.65 kg of Mg)*1.38kg of Mg
x kg of F=2.157Kg of F
2.157 kg of F*(10^3g/1kg )=2.16 x 10^3 g of F or 2,160 g of F
Answer contains 3 sig figs.
1.65 kg of Mg=2.58 kg of F
and
1.38 kg of Mg=x kg of F
So,
x kg of F/1.38kg of Mg=2.58 kg of F/1.65 kg of Mg
Solve for x kg of F,
x kg of F=(2.58 kg of F/1.65 kg of Mg)*1.38kg of Mg
x kg of F=2.157Kg of F
2.157 kg of F*(10^3g/1kg )=2.16 x 10^3 g of F or 2,160 g of F
Answer contains 3 sig figs.
1.98
2020