Thus we can balance the oxygen atoms by putting a prefix of 25/2 on the left side. To obtain a equation containing whole numbers, we multiply the entire equation by 2. This gives the final equation. 2 C8H18 + 25 O2 ---> 16 CO2 +18 H2O.
First you need to find out the Limiting reactant (LR). convert both reactants to the same thing. Check that the chemical equation is balanced. Now use stoichiometr and remember at moles, multiply: need moles, divide2 g / 42g/mol= 0.0477 mol propane mass propane/ Molar Mass propane = moles propane4 g / 32 g/mol= 0.125 mol oxygen X (1 mol/ 5 mol) = 0.025 mol propane oxygen is the LRmass O2 / MM O2 X (mol propane / mol O2)0.025 mol X (3 mol / 1 mol ) = .075 mol CO20.075 mol X (12 + 2*16) g /mol = 3.6 g CO2 In one step:2 g / 42g/mol X (3 mol / 1 mol ) X 48 g/mol = 6.86 g CO24 g / 32 g/mol X (3 mol / 5 mol ) X 48 g/mol = 3.6 g CO2mass/ MM X coefficient ratio X MM (new)
Answer:
32.4 mol
Explanation:
Given data:
Number of moles of C atom present = ?
Number of moles of glucose = 5.4 mol
Solution:
Glucose formula = C₆H₁₂O₆
There are 6 moles of C atoms are present in one mole of glucose.
In 5.4 moles of glucose:
5.4 mol × 6 = 32.4 mol
Spent fuel that can no longer be used to create energy is waste.
Answer: The pressure after the tire is heated to 17.3°C is 167 kPa
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
Putting values in above equation, we get:
Hence, the pressure after the tire is heated to 17.3°C is 167 kPa
