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:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
Answer:
2.765amu is the contribution of the X-19 isotope to the weighted average
Explanation:
The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:
X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21
The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:
19.00amu * 0.1455 =
2.765amu is the contribution of the X-19 isotope to the weighted average
