Question

3CaCO3(s) + 2H3PO49(aq) ---> Ca3(PO4)2(aq) + 3CO2(g) + 3 H20(l)
a)how many grams of phosphoric acid is needed to react with excess calcium carbonate to produce 3.74g of Ca3(PO4)2?

B)Calculate the number of grams of CO2 formed when 0.773g H20 is produced

Answer 1

phosphoric acid H3PO4

Answer 2

phosphoric acid is H3PO4

Answer 3

since it will produce 3.74g of Ca3(PO4)2, we'll use this to find the grams of phosphoric acid

Answer 4

\[3.74g Ca_3(PO_4)_2 \times \frac{1molCa_3(PO_4)_2}{310gCa_3(PO_4)_2} \times \frac{2molH_3PO_4}{1molCa_3(PO_4)_2} \times \frac{98gH_3PO_4}{1molH_3PO_4}\]

Answer 5

2.36g H3PO4, so do u understand it?

Answer 6

I don't really understand how you get there like why you do what you do. I understand what you do tho just not why

Answer 7

For the 1/310 is for converting Ca3(PO4)2 to moles, we divide by its molar mass, then for the middle part, which is 2/1, where moles of H3PO4 and Ca3(PO4)2 divides, shows that to convert the moles of Ca3(PO4)2 into moles H3PO4,

(remember that we cant straight away convert grams from Ca3(PO4)2 to grams of H3PO4, that's wrong, we can ONLY convert with moles to moles between elements)

For the third part, 98/1 indicates that to convert moles of H3PO4 into grams, so we multiply its molar mass.

Answer 8

Ohhhhhhh that makes sense!! how do you do B tho?

Answer 9

I'll assume that H2O is limiting, use back the same technique

Answer 10

0.773 x 1/18 x 3/3 x 44/1 = 1.89g CO2 do you agree?

Answer 11

....why?

Answer 12

same thing,
( 0.773 gH2O ) x (1molH2O/18gH2O ) x (3molCO2/3molH2O) x (44gCO2/1molCO2) = 1.89gCO2

Answer 13

Ok, makes sense now thanks so much