Condensation would occur at that time
Answer: 13.888g
Explanation:
CaCO3 —> CaO + CO2
From the equation above, 1mole of CaCO3 produces 1mole of CaO. Since the answer is to be expressed in grams, let us covert this mole to grams. This is illustrated below:
MM of CaCO3 = 40 +12 + (16x3) = 40 + 12 + 48 = 100g/mol
MM of CaO = 40 + 16 = 56g/mol
From the equation,
100g of CaCO3 produced 56g CaO
Therefore, 24.8g of CaCO3 will produce Xg of CaO i.e
Xg of CaO = (24.8 x 56) / 100 = 13.888g
Therefore, the theoretical yield of CaO is 13.888g
Answer:
154 g
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP
At STP, 1 mole of N₂ occupies 22.4 L.
79.5 L × 1 mol/22.4 L = 3.55 mol
Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂
The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.
Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃
The molar mass of NaN₃ is 65.01 g/mol.
2.37 mol × 65.01 g/mol = 154 g
Answer:
30 moles of water will produced.
Explanation:
Given data:
Number of moles of O₂ react = 15 mol
Number of moles of water formed = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Now we will compare the moles of water with oxygen.
O₂ : H₂O
1 : 2
15 : 2×15 = 30 mol
30 moles of water will produced.
Answer:
I think the answer is 6.022*10^23
