Balance the equation below to answer the next five questions. Ca_5(PO_4)_3 F + H
ID: 533444 • Letter: B
Question
Balance the equation below to answer the next five questions. Ca_5(PO_4)_3 F + H_2SO_4 rightarrow CaSO_4 + H_3PO_4 + HF How many grams of Ca_5(PO_4)_3F are necessary to produce 10.0 g of CaSO_4? a) 6.23 g b) 7.41 g c) 8.57 g d) 9.97 g e) 10.25 How much H_3PO_4 is produced (theoretically) from 10.0 g of H_2SO_4? a) 2.01 g b) 17.3 g c) 10.0 g d) 30.4 g e) 5.99 g Given that 80.0 g of Ca_5(PO_4)_3 F is reacted with 70.0 g of H_2SO_4, how much CaSO_4 is expected? a) 82.3 g b) 78.5 g c) 97.2 g d) 108 g e) 55.8 g How much of the reagent in excess in the previous problem is left over after the reaction is complete? a) 8.1 g b) 6.6 g c) 5.3 g d) 4.9 g e) 3.3 g If any 43.0 g of CaSO_4 was experimentally obtained from the reaction what is the percent yield of CaSO_4?Explanation / Answer
Ans:-1 The given equation is
Ca5(PO4)3F + 5H2SO4 -> 5CaSO4 + 3H3PO4 + HF
Now molar mass of Ca5(PO4)3F = 504.31g
molar mass of H2SO4 = 98g
molar mass of CaSO4 =136.14g
molar mass of H3PO4 = 97.994g
molar mass of HF = 20.01g
Now , Amount of Ca5(PO4)3F required to produce 10g of CaSO4
= 10g x 1x1x504.31 / 1x136.14x5x1
= 7.41g
Ans 2) Amount of H3PO4 produced from 10g H2SO4 = 10 x3x97.99 / 98x5
=5.99g
Ans 4) In the above question
80g of Ca5(PO4)3 F is reacted with 70g H2SO4
Now Moles of Ca5(PO4)3F =80 /504.31
=0.158
Moles of CaSO4 from Ca5(PO4)3F = 0.158x5
= 0.793
Now Moles of H2SO4 =70/98
=0.714
Moles of CaSO4 from H2SO4= 0.714x5/5
=0.714
Here Ca5(PO4)3F is the excess reagent
Therefore Amount of excess reagent left over after the reaction is complete
0.714/5 =0.1428mol Ca5(PO4)3F
Mass of Ca5(PO4)3F = 0.1428x504.31
=72.04g
Amount of excess Ca5(PO4)3F=80 - 72.04
=8.1g
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