<span> The </span>formula mass<span> is calculated by adding up all the atomic </span>masses <span>for every atom in the </span>formula<span>. First, we need to know the atomic masses of the atoms involved.
H = 1.01 amu
S = 32.06 amu
Formula mass = 2 x </span>1.01 amu + 2 x 32.06 amu = 66.14 formula mass unit
Answer:
The answer to your question is 6.0 moles of O₂
Explanation:
Data
2KClO₃ ⇒ 2KCl + 3O₂
moles of O₂ = ?
moles of KCl = 4
Process
To find the number of moles of O₂, use proportions and cross multiplication.
Use the coefficients of the balanced equation.
2 moles of KCl ----------------- 3 moles of O₂
4 moles of KCl ----------------- x
x = (4 x 3) / 2
-Simplification
x = 12/2
-Result
x = 6 moles of O₂
-Conclusion
When 4,0 moles of KCl are produced, 6.0 moles of O₂ will be produced.
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>
The answer is A. Solids only
