Question

A student is given a 3.589 g mixture of iron fillings, calcium chloride, & sand. The mixture is separated & the recoveries are as follows:

0.897 g of iron fillings

1.686 g of calcium chloride

0.923 g of sand

Calculate the percentage of sand:

Answer 1

Answer:

Mass of the mixture = 6.216 g

Mass of iron recover = 2.524 g

Percentage of iron recovered from the mixture:

\frac{2.524 g}{6.216 g}\times 100=40.60\%

6.216g

2.524g

×100=40.60%

Mass of sand recover = 1.932 g

Percentage of sand recovered from the mixture:

\frac{1.932 g}{6.216 g}\times 100=31.08\%

6.216g

1.932g

×100=31.08%

Mass of calcium chloride recover = 1.523 g

Percentage of calcium chloride recovered from the mixture:

\frac{1.523 g}{6.216 g}\times 100=24.50\%

6.216g

1.523g

×100=24.50%

Mass of mixture lost during separation :

6.216 g - 2.524 g - 1.932 g- 1.523 g = 0.237 g

Percent of material he lost during the separation process:

\frac{0.237 g}{6.216 g}\times 100=3.81\%

6.216g

0.237g

×100=3.81%

Answer 2

→ Question -:

A student is given a 3.589 g mixture of iron fillings, calcium chloride, & sand. The mixture is separated & the recoveries are as follows:

0.897 g of iron fillings

1.686 g of calcium chloride

0.923 g of sand

Calculate the percentage of sand:

→  Answer -:

Iron -:      2.524/6.216= 0.406= 40.6%

Sand -:    1.932/6.216= 0.311= 31.1%

NaCl -:     1.523/6.216= 0.245= 24.5%

Percent recovered -:   (2.524+1.923+1.523)/6.216= 0.96= 96%