How many grams of Sn3(PO4)2 are present in 0.45 moles of Sn3(PO4)2?

Question

Thanks and show your work!!!!!!!!!!!!

Answer 1

Hi Sophia, in order to calculate the grams of Sn₃(PO₄)₂, we need to calculate the molar mass of Sn₃(PO₄)₂. This can be done by looking at a periodic table and adding the molar masses of the elements together (taking into account how many of each element there is in the entire compound):

The molar mass of Sn is 118.71 g/mol. Because there are 3 x Sn in the compound, we would multiply this value by 3 to get 356.13 g

Next, we need to find the molar mass of P, which is 30.974 g/mol in the periodic table. Because there are 2 x P, this will be multiplied by 2 to get 61.948 g

Finally, we need to look at the molar mass of O, which is 15.999 g/mol. Because there are 2 x 4 x O, we will multiply this value by 8 to get 127.992 g.

The molar mass of Sn₃(PO₄)₂ will therefore by 356.13 g + 61.948 g + 127.992 g = 546.07 g/mol

Using this value, we can get the number of grams in 0.45 moles through a stoichiometric conversion:

0.45 moles of Sn₃(PO₄)₂ x (546.07 g/mol) = 245.73 g.

If you need to consider significant figures, because 0.45 moles has two sig figs, we will round our answer to also have two sig figs, and the answer with the correct sig figs would be 250 g. If you need a good periodic table online, I like to use ptable.com. Hope this is helpful!