Answer:
Explanation:
Hello,
In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:
- Moles of carbon are contained in the 9.582 grams of carbon dioxide:
- Moles of hydrogen are contained in the 3.922 grams of water:
- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:
Finally, we compute the percent by mass of oxygen:
Regards.
Regard the principle of utilization of two gas.
Make a consistent control of hardware containing gas.
Make a consistent control of weight diminishing valves giving gas.
No smoking zone.
Answer:
15.4 g of Zn₃(PO₄)₂ are produced
Explanation:
Given data:
Mass of zinc phosphate formed = ?
Volume of zinc nitrate = 48.1 mL (0.05 L)
Molarity of zinc nitrate = 2.18 M
Solution:
Chemical equation:
3Zn(NO₃)₂ + 2K₃PO₄ → Zn₃(PO₄)₂ + 6KNO₃
Moles of zinc nitrate:
Molarity = number of moles / volume in litter
Number of moles = 2.18 M × 0.05 L
Number of moles = 0.109 mol
Now we will compare the moles of zinc phosphate with zinc nitrate from balanced chemical equation:
Zn(NO₃)₂ : Zn₃(PO₄)₂
3 : 1
0.109 : 1/3×0.109 = 0.04 mol
0.04 moles of Zn₃(PO₄)₂ are produced.
Mass of Zn₃(PO₄)₂:
Mass = number of moles × molar mass
Mass = 0.04 mol × 386.1 g/mol
Mass = 15.4 g
Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:
It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:
the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
