Clean? I’m pretty sure not sure what it means but.
Answer:
46.40 g.
Explanation:
- It is a stichiometric problem.
- The balanced equation of the reaction: 4K + O₂ → 2K₂O.
- It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.
- We should convert the mass of K (38.5 g) into moles using the relation:
<em>n = mass / molar mass,</em>
n = (38.5 g) / (39.098 g/mol) = 0.985 mole.
<em>Using cross multiplication:</em>
4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.
0.985 mole of K produces → ??? moles of K₂O.
∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.
- Now, we can get the mass of K₂O:
∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.
Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
Answer:
The number of moles of benzaldehyde = 0.0253 moles
Explanation:
The molecular formula of benzaldehyde is C₇H₆O
Its molecular mass is calculated from the atomic masses of the constituent atoms.
C = 12.0 g: H = 1.0 g; O = 16.0 g
Molecular mass = ( 12 * 7) + (1 * 6) + (16 * 1) = 106.0 g/mol
Number of moles of substance = mass of substance/ molar mass of the substance
mass of benzaldehyde = 2.68; molar mass = 106.0 g/mol
Number of moles of benzaldehyde = 2.68 g/ 106 g/mol = 0.0253 moles
Therefore, the number of moles of benzaldehyde = 0.0253 moles
Answer:
Chemical changes cause a substance to change into an entire substance with a new chemical formula. Chemical changes are also known as chemical reactions. The “ingredients” of a reaction are called reactants, and the results are called products.
Hope it helps
