Answer:
[See Below]
Explanation:
I'd say 44 something. It's probably ml but I can't see what it says on the tube.
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>
<span>Pre-1982 definition of STP: 37 g/mol
Post-1982 definition of STP: 38 g/mol
This problem is somewhat ambiguous because the definition of STP changed in 1982. Prior to 1982, the definition was 273.15 K at a pressure of 1 atmosphere (101325 Pascals). Since 1982, the definition is 273.15 K at a pressure of exactly 100000 Pascals). Because of those 2 different definitions, the volume of 1 mole of gas is either 22.414 Liters (pre 1982 definition), or 22.71098 liters (post 1982 definition). And finally, there's entirely too many text books out there that still use the 35 year obsolete definition. So let's solve this problem using both definitions and you need to pick the correct answer for the text book you're using.
First, determine how many moles of gas you have. Just simply divide the volume you have by the molar volume.
Pre-1982: 2.1 / 22.414 = 0.093691443 moles
Post-1982: 2.1 / 22.71098 = 0.092466287 moles
Now determine the molar mass. Simply divide the mass by the moles. So
Pre-1982: 3.5 g / 0.093691443 moles = 37.35666667 g/mol
Post-1982: 3.5 g / 0.092466287 moles = 37.85163333 g/mol
Finally, round to 2 significant figures. So
Pre-1982: 37 g/mol
Post-1982: 38 g/mol</span>
Answer:
34 gram of FeO produced 8 gram of oxygen.
Explanation:
Given data:
Mass of FeO = 34 g
Mass of oxygen = ?
Solution;
Chemical equation:
2FeO → 2Fe + O₂
Number of moles of FeO:
Number of moles = mass/ molar mass
Number of moles = 34 g /71.8 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of FeO with oxygen:
FeO : O₂
2 : 1
0.5 : 1/2 × 0.5 = 0.25
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 0.25 mol × 32 g/mol
Mass = 8 g
So 34 gram of FeO produced 8 gram of oxygen.
