Answer:
1610.7 g is the weigh for 4.64×10²⁴ atoms of Bi
Explanation:
Let's do the required conversions:
1 mol of atoms has 6.02×10²³ atoms
Bi → 1 mol of bismuth weighs 208.98 grams
Let's do the rules of three:
6.02×10²³ atoms are the amount of 1 mol of Bi
4.64×10²⁴ atoms are contained in (4.64×10²⁴ . 1) /6.02×10²³ = 7.71 moles
1 mol of Bi weighs 208.98 g
7.71 moles of Bi must weigh (7.71 . 208.98 ) /1 = 1610.7 g
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:
