Answer:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
Answer:
280.8 g
Explanation:
Definimos la reaccion:
2NaOH + FeSO₄ → Na₂SO₄ + Fe(OH)₂
Como tenemos la masa de NaOH, asumimos que el sulfato de hierro (II) es el reactivo en exceso.
Definimos masa de reactivo: 250 g . 1mol / 40g = 6.25 mol
2 moles de NaOH producen 1 mol de hidroxido ferroso
Entonces 6.25 moles producirán, la mitad (6.25 . 1) /2 = 3.125 moles
Convertimos los moles a masa:
3.125 mol . 89.85 g/mol = 280.8 g
