Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.
We make a proportion out of the word problem
(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose
Answer:
See which of your answer choices are comparable and what you can contrats them with
Explanation:
YEET
Answer: 323.61 g of 
will be produced
Explanation:
The given balanced chemical reaction is :
According to stoichiometry :
2 moles of 
require 1 mole of 
Thus 3.00 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product.
As 2 moles of give = 2 moles of
Thus 3.00 moles of give =
of
Mass of 
Thus 323.61 g of will be produced from the given moles of both reactants.
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
