Answer:
Mass = 182.4 g
Explanation:
Given data:
Number of moles of Al₂O₃ = 3.80 mol
Mass of oxygen required = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Now we will compare the moles of aluminum oxide and oxygen.
Al₂O₃ : O₂
2 : 3
3.80 : 3/2×3.80 = 5.7
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 5.7 mol × 32 g/mol
Mass = 182.4 g
Answer:
Explanation: When solutions of potassium iodide and lead nitrate are combined?
The lead nitrate solution contains particles (ions) of lead, and the potassium iodide solution contains particles of iodide. When the solutions mix, the lead particles and iodide particles combine and create two new compounds, a yellow solid called lead iodide and a white solid called potassium nitrate. Chemical Equation Balancer Pb(NO3)2 + KI = KNO3 + PbI2. Potassium iodide and lead(II) nitrate are combined and undergo a double replacement reaction. Potassium iodide reacts with lead(II) nitrate and produces lead(II) iodide and potassium nitrate. Potassium nitrate is water soluble. The reaction is an example of a metathesis reaction, which involves the exchange of ions between the Pb(NO3)2 and KI. The Pb+2 ends up going after the I- resulting in the formation of PbI2, and the K+ ends up combining with the NO3- forming KNO3. NO3- All nitrates are soluble. ... (Many acid phosphates are soluble.)
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g
