What mass of P4 is possible from the reaction of 45.0g of Ca3(PO4)2, 70.0 of SiO2 and 18.0g of C? The balanced reaction is: 2 Ca3(PO4)2 (s) + 6 SiO2 (s) + 10 C(s) ---> P4 (s) + 6 CaSiO3 (l)+ 10 CO(g)
What mass of P4 is possible from the reaction of 45.0g of Ca3(PO4)2, 70.0 of SiO2...
What is the maximum mass of P4 that can be obtained from 56.2 g Ca3(PO4)2, 37.3 g SiO2 and excess carbon? The chemical equation for the reaction is given below. 2 Ca3(PO4)2(s) + 6 SiO2 + 10 C(s) ⟶ P4(g) + 6 CaSiO3(l) + 10 CO(g) Give your answer in grams with three significant figures. Molar masses, in g mol-1: Ca3(PO4)2, 310.18 SiO2, 60.09 P4, 123.88
DuPPY U111204) 20. Elemental phosphorous, P4, can be prepared by heating calcium phosphate, Ca3(PO4)2, with sand (silicon dioxide, SiO2) and coke (impure C). The balanced equation for this reaction is shown in Equation 18. d 2 Ca3(PO4), (s) + 6 SiO2(s) + 10C(s) + P (g) + 6 CaSiO3(s) + 10 CO(g) (Eq. 18) o What is th What is the percent yield of P, if 27.3 g of P4 are recovered from the reaction of 200 g of Ca3(PO4)2...
Elemental phosphorus is produced by the reaction, 2Ca3(PO4)2 +6SiO2 +10C→6CaSiO3 +10CO+P4 Suppose that you have 7.00 moles of Ca3(PO4)2, 23.0 moles of SiO2, and 42.0 moles of C. (a) Which reactant is limiting? (b) What are the maximum amounts (in moles) of CaSiO3, CO, and P4 that can be produced from these amounts of reactants?
Phosphorus can be prepared from calcium phosphate by the following reaction: 2 Cas (POA),(s) + 6 SiO2 (s) + 10C(s) + 6 CaSiO2 (s) + P4(s) + 10 CO(9) Phosphorite is a mineral that contains Ca3(PO4), plus other non-phosphorus-containing compounds. What is the maximum amount of P4 that can be produced from 1.9 kg of phosphorite if the phosphorite sample is 75% Ca3(PO4), by mass? Assume an excess of the other reactants. Mass =
Write a balanced equation for the following by inserting the correct coefficients (if a coefficient = 1, enter "1" in the provided box). C(s)+ Ca3(PO4)2(s) + SiO2(s)→1P4(g) + CO(g) + CaSiO3(l)
Phosphoric acid can be prepared from calcium phosphate according to the following reaction: Ca3(PO4)2 (s) + 3 H2SO4 (l) → 3 CaSO4 (s) + 2 H3PO4 (l) 310. g/mol 98.0 g/mol 136 g/mol 98.0 g/mol [a] If 0.664 mole of Ca3(PO4)2 are combined with 1.53 mole of H2SO4, how many grams of H3PO4 can be produced? [4 pts] [b] If 62.6 g of H3PO4 is actually produced in the above reaction, what is the percent yield for the reaction? [2...
The Ksp of Ca3(PO4)2 is 1x 10-33. What is the solubility in moles per liter (mol/L) of Ca3(PO4)2 (s) in a 0.01 M Ca(NO3)2 (aq) solution? (Note: The molar mass of Ca3(PO4)2 is not needed in this problem, which asks for moles per L.) a. 3.2 x 10-14 mol/L b.3.0 x 10-15 mol/L c. 9.8 x 10 mol/L d. 1.6 x 10-14 mol/L e. 5.0 x 10-32 mol/L
---Percent yield of Ca3(PO4)2 from CaCL2 Mass of beaker + CaCL2 = 342.231g, Mass of beaker = 341.852g, Mass of filter paper + Ca3(PO4)2 ppt = 0.726, Mass of filter paper = 0.506 1. Show the calculation for theoretical yield of Ca3(PO4)2 2. Show the calculation for percent yield of Ca3(PO4)2
The solubility of Ca3(PO4)2 (molar mass = 310 g/mol) in water is 0.065 g/L. The Ksp for Ca3(PO4)2 is ______ (scientific notation with 2 sig fig, such as 3.5e-9)
What is the concentration of calcium ion that results from the unbalanced reaction Ca3(PO4)2(s)= Ca2+ (aq) +P04(ag) Where K=1.77x10-4