Answers
Answer:
The number of moles of H₃PO₄ that reacted is 0.000343 moles
Note: Some data is missing. Data from the attachment is used for the calculationsinnthe explanation below.
Explanation:
The reaction is a neutralization reaction between NaOH and H₃PO₄. The equation of the reaction is given as follows:
3 NaOH + H₃PO₄ ---> Na₃PO₄ + 3 H₂O
The molarity of the NaOH solution is 0.238 mol/L.
Average volume of NaOH used during the titration to arrive to endpoint = (4.6 + 3.9 + 4.5) mL / 3 = 4.33 mL
Molarity is defined ratio of the number of moles of solute to the volume of solution. Mathematically, molarity = number of moles/volume in Litres
Number of moles of NaOH reacted = 0.238 mol/L × (4.33mL × 1 L/1000 mL)
Number of moles of NaOH = 0.00103 moles
From the equation of the reaction, 3 moles of NaOH reacts with 1 mole of H₃PO₄
0.00103 moles of NaOH will react with 0.00103 x 1/3 moles of H₃PO₄ = 0.000343 moles of H₃PO₄.
Therefore, number of moles of H₃PO₄ that reacted is 0.000343 moles
Related Questions
En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción
Answers
Answer:
88.9%
Explanation:
Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:
5.97 g ÷ 180 g/mol = 0.0332 molDespués calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:
0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:
0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 molFinalmente calculamos el rendimiento porcentual:
0.0590 mol / 0.0664 mol * 100% = 88.9%Which equation obeys the law of conservation of
mass?
Answers
Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
Que es la actividad física y en qué mejora
Answers
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answers
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
