Answer:
1.48 moles of SeCl6 are needed
Explanation:
Based on the reaction:
SeCl6 + O2 → SeO2 + 3Cl2
<em>1 mole of SeCl6 reacts producing 3 moles of Cl2.</em>
To solve this question we need to use the conversion factor:
1mol SeCl6 = 3mol Cl2
As we want to produce 4.45 moles of Cl2, we need:
4.45 mol Cl2 * (1mol SeCl6 / 3mol Cl2) =
<h3>1.48 moles of SeCl6 are needed</h3>
1. Translate, predict the products, and balance the equation above.
Li + Cu(NO3)2 = Li(NO3)2 + Cu
2. How many particles of lithium are needed to produce 125 g of copper?
125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles
3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?
4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = <span>1043.77 grams Li(NO3)2</span>
The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>
Answer:
Moles de LiOH= 0.33212 Moles
Explanation:
1 Mol of LiOH -->6.022*10^23 molecules of LiOH
therefore
