C₄H₉OH + HBr = C₄H₉Br + H2O
Δmole of alcohol gives 1 mole of bromobutanol
HBr is in excess, so the yield of the product is limited by the alcohol
Wt. of 1 butanol = 18
Molar mass of the butanol = 74.12 g/mole
Moles of the alcohol = 1/74.12 = 0.01349 moles
So, moles of bromobutane = 0.01349 moles
Molar mass of C₄H₉Br = 137.018 g/moles
So, theoretical mass of bromobutane is = 0.01349 × 137.0.18
= 1.85 g
The balanced chemical reaction:
<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
Answer:
600 mAh·g−1
Explanation:
i hope this is good let me know if its wrong
Answer:
0.665 moles of CO₂
Explanation:
The balance chemical equation for the combustion of Ethane is as follow:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
Step 1: <u>Calculate moles of C₂H₆ as;</u>
Moles = Mass / M.Mass
Putting values,
Moles = 10.0 g / 30.07 g/mol
Moles = 0.3325 moles
Step 2: <u>Calculate Moles of CO₂ as;</u>
According to balance chemical equation,
2 moles of C₂H₆ produced = 4 moles of CO₂
So,
0.3325 moles of C₂H₆ will produce = X moles of CO₂
Solving for X,
X = 0.3325 mol × 4 mol ÷ 2 mol
X = 0.665 moles of CO₂
