A 5.00 mL sample of an aqueous solution of H3PO4 requires 55.0 mL of 0.258 M NaOH to convert all of the H3PO4 to Na2HPO4. moles P4O10 = mass / molar mass = 10.00 g / 283.88 g/mol = 0.035226 mol. Example Reactions: • PH3 + 2 O2 = H3PO4 • 3 KOH + H3PO4 = K3PO4 + 3 H2O • H3PO4 + 5 HCl = PCl5 + 4 H2O • H3PO4 + 3 NH3 = (NH4)3PO4 • H3PO4 + NaOH = NaH2PO4 + H2O The other product of the reaction is water. Calculations: Formula: H3PO4 Molar Mass: 97.9937 g/mol 1g=1.02047376515021E-02 mol Percent composition (by mass): Element Count Atom Mass %(by mass) Mass Na3PO4 formed = 0.833 mol X 163.9 g/mol = 137 g Na3PO4 H3PO4 is a colorless viscous liquid at room temperature. 1. Solution The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the elements in H3PO4 is calculated as follows: %H=(3(1.008 g) H/ 97.99 g H3PO4)*100%= 3.086% mole H3PO4 = 1430 g H3PO4 x 1 mole H3PO4/98 g H3PO4 = 15 moles. b. d. How many H atoms are there in 2.5 mol H3PO4.2.5pt Now there is 15 moles of H3PO4 in the initial liter so it is 15M H3PO4! Do a quick conversion: 1 grams H3PO4 = 0.010204583427424 mole using the molecular weight calculator and the molar mass of H3PO4. Molar Mass: 97.9952. Molar mass of *2H3PO4 is 195.9904 g/mol Convert between *2H3PO4 weight and moles From its MSDS sheet it is 1.680g/mL. - 13159451 ams) has entries 4.9, 14.7. How many oxygen atoms are in one unit of H3PO4? Complete the following calculations for H3PO4.4 pts (1pt each) a. P4O10 + 6H2O ----> 4 H3PO4. See also our theoretical yield calculator for chemical reactions (probably your next stop to finish the problem set). What is the molar mass of H3PO4? Mass H3PO4 remaining = 18.3 g H3PO4 remaining. Start with a liter of solution which is 1000 mLx 1.680 g/mL gives a mass of 1680 g. mass H3PO4 = 1680 g solution x 85 parts H3PO4/100 parts solution = 1430 g H3PO4. work out the limiting reagent. The one that produces the least H3PO4 is the limiting reagent. To do this work out how much H3PO4 you would get if all of each reagent individually were to fully react. What am I doing wrong? c. How many moles of PO43-ions are present in 2.5 moles of phosphoric acid, H3PO4? I don't know what to do with this, I tried: (.005 L)(x M) = (.055 L)(0.258 M) And I got 2.838 mol/L But the answer is wrong apparently. moles H2O = 12.0 g / 18.016 g/mol = 0.66607 mol. Mass H3PO4 consumed = 2.50 mol NaOH X (1 mol H3PO4 / 3 mol NaOH) X 97.99 g/mol = 81.7 g H3PO4 consumed. Calculate the formula and molar mass. molar mass of H3PO4 IS 97.99g/mol. The third column labeled Mass of Washers m subscript w (kilograms) has entries 0.0049, empty. moles of H3PO4 = 25.4 g x a million mol/ninety 8 g = 0.259 mol H3PO4 Kg of solvent = (100g - 25.4 g) = seventy 4.6 g = 0.0746 kg Molality = moles of solute/kg of solvent = 0.259 mol/0.0746 kg solvent Molality = 3.47m Can someone give me a step by step to the solution? 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