Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
The reaction will produce 12.1 g Ag₂S.
<em>Balanced equation</em> = 2Ag + S ⟶ Ag₂S
<em>Mass of Ag₂S</em> = 10.5 g Ag × (1 mol Ag/107.87 g Ag) × (1 mol Ag₂S/2 mol Ag)
× (247.80 g Ag₂S/1 mol Ag₂S) = 12.1 g Ag₂S
Answer:
1 mole of HCl or NaOH gives you 1 mole of H2O , then the number of moles in H2O is: [ 1÷1×1 ] = 1 mole.
Explanation:
Answer:
34 gram of FeO produced 8 gram of oxygen.
Explanation:
Given data:
Mass of FeO = 34 g
Mass of oxygen = ?
Solution;
Chemical equation:
2FeO → 2Fe + O₂
Number of moles of FeO:
Number of moles = mass/ molar mass
Number of moles = 34 g /71.8 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of FeO with oxygen:
FeO : O₂
2 : 1
0.5 : 1/2 × 0.5 = 0.25
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 0.25 mol × 32 g/mol
Mass = 8 g
So 34 gram of FeO produced 8 gram of oxygen.
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