Answer:
(edit: nvm I figured it out, here is the answer)
Explanation:
Answer:
2.58 L
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
AsF3:C2CI6
4:3
1.3618 moles: 1.02135 moles(1.3618÷4×3)
C2CI6 is the limting reagent
So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)
or
Balanced equation
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.
Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.
Explanation:
This works because it demonstrates that as volume increases, pressure decreases (inverse relationship)
Answer:
Write a balanced chemical reaction:
N2 + 3H2 ==> 2NH3
Looking at the mole ratios in this balanced equation you can see it takes 3 moles H2 to make 2 moles NH3. So, next calculate the moles of NH3 represented by 1.80 g and then convert to moles of H2 needed:
moles of NH3 = 1.80 g x 1 mole/17 g = 0.106 moles NH3
Moles H2 needed = 0.106 moles NH3 x 3 moles H2/2 moles NH3 = 0.159 moles H2 needed
Grams H2 needed = 0.159 moles x 2 g/mole = 0.318 grams H2 needed
Explanation:
