Molar mass O2 = 31.99 g/mol
Molar mass CO2 = 44.01 g/mol
Moles ratio:
<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2
mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01
mass of O2 = 8174.8575 / 132.03
mass of O2 = 61.916 g
Therefore:
1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916
moles O2 = 61.916 x 1 / 31.99
moles = 61.916 / 31.99 => 1.935 moles of O2
Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
- 15 molecules H₂ *
= 10 molecules NH₃
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.
Answer:
0.719M AgNO₃
Explanation:
Based on the reaction:
MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂
<em>1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃</em>
<em />
To find molarity of AgNO₃ solution we need to determine moles of AgNO₃ and, as molarity is the ratio of moles over liter (13.9mL = 0.0139L). Now, to determine moles of AgNO₃ we need to use the reaction, thus:
<em>Moles AgNO₃:</em>
<em />
Moles of MgBr₂ are:
50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.
As the silver nitrate reacts completely and 2 moles of AgNO₃ reacts per mole of MgBr₂:
0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =
0.0100 moles of AgNO₃ are in the solution.
And molarity is:
0.0100 moles AgNO₃ / 0.0139L =
<h3>0.719M AgNO₃</h3>
30kg think of it like this imagine you are moving to your college dorm would you have you carry the box if weights or will you carry your books instead
