The balanced combustion reaction of propane, C₃H₈, is
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane
Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen
Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen
Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.
Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams
Answer: Option (c) is the correct answer.
Explanation:
A limiting reagent is defined as a reagent that completely gets consumed in a chemical reaction. A limiting reagent limits the formation of products.
For example, we have given 5 mol of A and the reaction is 
Whereas when 4 mol B will react with 2 mol of A. Hence, 8 mol of B will react with 4 mol A as follows.
= 4 mol
As, the given moles of A is more than the required moles. Thus, it is considered as an excess reagent.
Hence, B is a limiting reagent because it limits the formation of products.
Thus, we can conclude that limiting reactant is the term used to describe the reactant that is used up completely and controls the amount of product that can be produced during a chemical reaction.
The total pressure = 1.402 atm
<u><em>calculation</em></u>
Total pressure = partial pressure of gas A + partial pressure of gas B + partial pressure of third gas
partial pressure of gas A= 0.205 atm
Partial pressure of gas B =0.658 atm
partial pressure for third gas is calculated using ideal gas equation
that is PV=nRT where,
p(pressure)=? atm
V(volume) = 8.65 L
n(moles)= 0.200 moles
R(gas constant)=0.0821 L.atm/mol.k
T(temperature) = 11°c into kelvin =11+273 =284 k
make p the subject of the formula by diving both side by V
p =nRT/v
p = [(0.200 moles x 0.0821 L.atm/mol.K x 284 K)/8.65L)] =0.539 atm
Total pressure is therefore = 0.205 atm +0.658 atm +0.539 atm
=1.402 atm
The answer to this is B, solids with a repeating atomic pattern.
Answer:
13.5 moles of AgNO₃
Explanation:
To determine the reaction:
Reactants: AgNO₃ and Cu
Products: Cu(NO₃)₂ and Ag
2 moles of AgNO₃ react to 1 mol of Cu, in order to produce 1 mol of Cu(NO₃)₂ and 2 moles of solid silver.
2AgNO₃ + Cu → Cu(NO₃)₂ + 2Ag
Our production was 6.75 moles of Cu(NO₃)₂
Let's make the rule of three:
1 mol of Cu(NO₃)₂ is produced by 2 moles of AgNO₃
Then, our 6.75 moles were definetely produced by (6.75 . 2) /1 = 13.5 moles.
If the copper was in excess, then the silver nitrate is the limiting reactant:
2 mol of AgNO₃ can produce 1 mol of Cu(NO₃)₂
Then, 13.75 moles of silver nitrate must produce (13.5 . 1) /2 = 6.75 moles of Cu(NO₃)₂
