Answer:
50.00 g of NO
Explanation:
Remember that the balanced chemical reaction equation is indispensable in solving any question that has to do with stoichiometry. Hence the first step in solving the problem is noting down the balanced chemical reaction equation.
2NO(g) + O2 (g)→ 2NO2(g)
Now we try to find out the reactant in excess. The reactant in excess gives the greater mass of product.
For O2;
From the balanced reaction equation;
32 g of O2 yields 92g of NO2
16.00g of O2 will yield 16.00×92/32 = 46g of NO2
For NO;
30g of NO yields 92g of NO2
80.00 g of NO yields 80.00 × 92/30 = 245.33 g of NO2
Hence NO is the reactant in excess.
If 1 mole of O2 reacts with 2 moles of NO2 according to the balanced reaction equation
Then 32 g of O2 reacts with 60g of NO according to the balanced reaction equation
16.00 g of O2 reacts with 16.00 × 60 /32 = 30 g of NO
Hence mass of excess reactant used in the reaction = total mass of NO- mass of NO reacted= 80.00g -30.00g = 50.00 g of NO
Hence the mass of excess reactant used in the reaction is 50.00 g of NO
Answer:
18 g
Explanation:
Step 1: Write the balanced equation
2 Al + 6 HCl ⇒ 2 AlCl₃ + 3 H₂
Step 2: Calculate the moles of Al
The molar mass of Al is 26.98 g/mol.
28.1 g × 1 mol/26.98 g = 1.04 mol
Step 3: Calculate the moles of HCl
46 mL of 1 M HCl react.
0.046 L × 1 mol/L = 0.046 mol
Step 4: Determine the limiting reactant
The theoretical molar ratio of Al to HCl is 2:6 = 0.33:1.
The real molar ratio of Al to HCl is 1.04:0.046 = 22.6:1
According to this, the limiting reactant is HCl.
Step 5: Calculate the mass of AlCl₃ generated from 0.046 moles of HCl
The molar ratio of HCl to AlCl₃ is 6:2. The molar mass of AlCl₃ is 133.34 g/mol.
0.046 mol HCl × 6 mol HCl/2 mol AlCl₃ × 133.34 g AlCl₃/1 mol AlCl₃ = 18 g AlCl₃
<em><u>Larger molecules take longer to move up the chromatography paper or TLC plate, whereas smaller molecules are more mobile. Likewise, the polarity of the molecules can affect how far the spots travel, depending on the type of solvent used.</u></em>
Answer: 2 moles of CO₂ are produced from reacting 2 moles of H₂CO with oxygen.
Explanation:
1) Chemical equation:
Formaldehyde + oxgyen → CO₂ + H₂O
H₂CO + O₂ → CO₂ + H₂O
2) That equation is already balanced
3) Mole ratios
1 mol H₂CO : 1 mol O₂ : 1 mol CO₂ : 1 mol H₂O
4) Then, every mol of mol of H₂CO produces 1 mol of CO₂ , which means that 2 moles of CO₂ are produced from reacting 2 moles of H₂CO with oxygen.