<u>Answer</u>
So this is the reaction that happens.
<span>C4H10 + O2 = CO2 + H2O </span>
<span>Balanced, it is </span>
<span>2C4H10 + 8O2 = 8CO2 + 10H2O </span>
<span>Given 1 kg or 1000 g of butane, use stoichiometry aka factor labeling aka conversions and mole ratios to get to grams of oxygen. </span>
<span>I'll do an example. Let's form water. Hydrogen is diatomic too. </span>
<span>2H2 + O2 = 2H2O </span>
<span>Given 1000 g of Hydrogen, I need to know how many grams of oxygen to use. To convert grams to moles,
I know that 1 mol of H2 equals 2.02 g. Then, for every mole of O2, there are 2 moles of H2. Then converting moles of O2 to grams, I know that one mole of it equals 32 grams. </span>
<span>[1000 g H2] x [1 mol H2/2.02 g H2] x [1 mol O2/2 mol H2] x [32 g O2/1 mol O2] </span>
<span>My answer would be 7.9 kg </span>
The hallogens chloride with br
Answer:
you can solve the rest of the equation. I only reduced it to that much to show you how to derive it
Answer:
81.26% is the percent yield
Explanation:
Based on the reaction:
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>
<em />
To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:
Actual yield (0.366g) / Theoretical yield * 100
<em>Moles CaCl₂ = Moles CaCO₃:</em>
0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃
<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>
0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃
Percent yield = 0.366g / 0.450g * 100
81.26% is the percent yield
Answer:
Neils Bohr determined that electrons inhabit distinct energy levels.
Explanation:
