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:
B
Explanation:
The four factors that increase the rate of the reaction:
1.Concentration of the reactants
2.Size of the particles
3.Temperature
4.Catalyst
Calcium Phospate formula: Ca3(PO4)2
Atomic weight:
Ca = 40 ; P = 31 ; O = 16
Ca3 = 40 * 3 = 120
P = 31
O4 = 16 * 4 = 64
(PO4)2 = (31 + 64) * 2 = 95 * 2 = 190
Ca3(PO4)2 = 120 + 190 = 310 g/mol
31/310 = 10% P in the calcium phospate
1.00kg * 1000 g/kg = 1000 g of phosphorus.
1000 g / 10% = 10,000 g of calcium phosphate
10,000 g / 58.5% = 17,094 grams of ore.
The minimum mass of the ore should be 17,094 grams or 17.094 kg.
Try this solution:
if 1 mole is 55.8 gr., then 2.25 moles the mass should be 55.8*2.25=125.55 gr.
Answer: 125.55 gr.
