Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).
Answer:
5.7*10^4 is equal to 57,000.
Explanation:
First, we must multiply 10 by its power, 4. That would be 10 4 times.
10*10*10*10 = 10,000.
Then, we multiply it by 5.7.
5.7*10,000 = 57,000.
Regards!
You can view more details on each measurement unit: molecular weight of Aluminium or grams The molecular formula for Aluminium is Al. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Aluminium, or 26.981538 grams.
Answer:
Explanation:
<u>1) Balanced chemical equation:</u>
<u>2) Mole ratio:</u>
- 2 mol S : 3 mol O₂ : 2 mol SO₃
<u>3) Limiting reactant:</u>
n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂
n = 7.0 g / 32.065 g/mol = 0.2183 mol S
Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859
Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5
Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.
<u>4) Calcuate theoretical yield (using the limiting reactant):</u>
- 0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃
- x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃
<u>5) Yield in grams:</u>
- mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g
<u>6) </u><em><u>Percent yield:</u></em>
- Percent yield, % = (actual yield / theoretical yield) × 100
- % = (7.9 g / 10.0 g) × 100 = 79%