From the coefficients of the equation, we know that for every 3 moles of water consumed, 1 mole of diphosphorus trioxide is consumed.
This means we need to find the mass of 0.75 moles of diphosphorus trioxide.
- The atomic mass of phosphorous is 30.973761998 g/mol.
- The atomic mass of oxygen is 15.9994 g/mol.
So, the formula mass of diphosphorus trioxide is:
- 2(30.973761998)+3(15.9994)=109.945723996 g/mol.
Thus, 0.75 moles have a mass of:
- 0.75(109.945723996), which is about 82.5 g (to 3 sf)
Answer:
1326.6 × 10⁻²³g
Explanation:
Given data:
Number of atoms of iron = 143 atoms
Average atomic mass of iron = 55.845 amu
Mass of iron = ?
Solution:
It is known that,
55.845 g of iron = one mole of iron = 6.02× 10²³ atoms
For 143 atoms:
143 atoms /6.02× 10²³ atoms × 55.845 g
23.754 × 10⁻²³× 55.845 g
1326.6 × 10⁻²³g
Answer:
C
Explanation:
with combustion you must know that your products must be carbon dioxide and water. I think you copied your question incorrectly.
Answer:
Mass of cisplatin = 72.32 g
Explanation:
Given data:
Mass of KPtCl = 100 g
Mass of Cisplatin = ?
Solution:
First of all we will write the balance chemical equation.
Chemical equation:
K₂PtCl₄ (aq) + 2NH₃(aq) → Pt(NH₃)₂Cl₂(s) + 2KCl (aq)
Number of moles of = mass / molar mass
Molar mass of K₂PtCl₄ = 39.0983 ×2 + 195.084 + 35.5× 4
Molar mass of K₂PtCl₄ = 415.284 g/mol
Number of moles of K₂PtCl₄ = 100 g / 415.284 g/mol
Number of moles of K₂PtCl₄ = 0.241 mol
Now we will compare the moles of K₂PtCl₄ with Cisplatin from balance chemical equation.
K₂PtCl₄ : Pt(NH₃)₂Cl₂
1 : 1
0.241 : 0.241
Ammonia is in excess so mass of cisplatin depend upon the K₂PtCl₄.
Mass of cisplatin:
Mass of cisplatin = Number of moles × Molar mass
Mass of cisplatin = 0.241 mol × 300.1 g/mol
Mass of cisplatin = 72.32 g
