2C3H7OH + 9O2 → 6CO2 + 8H2O
Answer:
24.5 g of NaCl
Explanation:
We begin from the balanced reaction:
3MgCl₂ + 2Na₃PO₄ → 6NaCl + Mg₃(PO₄)₂
If the sodium phosphate is in excess, then the limting reagent is the magnessium chloride.
We convert mass to moles:
20 g . 1mol / 95.2g = 0.210 moles.
3 moles of MgCl₂ can produce 6 moles of NaCl
0.210 moles of salt, may produce (0.210 . 6) /3 = 0.420 moles
Ratio of reactant is twice the product
We convert the moles to mass:
0.420 mol . 58.45 g/mol = 24.5 g
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
The answer to this ? is true
Answer:
2.14 moles of H₂O₂ are required
Explanation:
Given data:
Number of moles of H₂O₂ required = ?
Number of moles of N₂H₄ available = 1.07 mol
Solution:
Chemical equation:
N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
now we will compare the moles of H₂O₂ and N₂H₄
N₂H₄ : H₂O₂
1 : 2
1.07 : 2×1.07 = 2.14 mol
