Question

4. Given that 4NH3 + 5O2 4NO + 6H2O, if 4.23 x 1022 molecules NH3 were made to react with an excess of oxygen gas, how many molecules of NO would form?
5. If 0.433 moles of sulfur react with 0.500 moles of chlorine, how many moles of disulfur dichloride are produced? Which reactant is the limiting reactant and which is the reactant in excess?
S8(l) + 4Cl2(g)  4S2Cl2(l)


6. How many grams of Fe2O3 are produced when 2.30x10^24 molecules of O2 are reacted?

7. 3.50 g of potassium reacts with water to produce potassium hydroxide and hydrogen gas.  Calculate the mass of potassium hydroxide produced.  The unbalanced equation is:  K + H2O KOH + H2

Please help as much as possible!

Answer 1

Answer:

Explanation:

Given Rxn =>  4NH₃ + 5O₂ => 4NO + 6H₂O

Given data => 4.23 x 10²² molecules NH₃ => ? molecules NO

Approach: Convert given value in molecules to moles, solve for moles NO by equation ratios in balanced equation. Finish by multiplying moles of NO by Avogadro's Number (= 6.02 x 10²³ molecules/mole) to obtain molecules of NO.

moles NH₃ = 4.23 x 10²² molecules NH₃ / 6.023 x 10²³ molecules/mole

= 0.0703 mole NH₃

From equation stoichiometry of balanced equation 4 moles NH₃ gives 4 moles NO. Then 0.0703 mole NH₃ => 0.0703 mole NO b/c coefficients are equal in balanced equation.

∴molecules of NO = 0.0703 mole NO x 6.03 x 10²³ molecules NO/mole NO

= 4.23 x 10²² molecules of NO.

The remaining problems can be worked much in the same way. Convert given data to moles (if not already expressed in terms of moles), apply equation ratios to calculate needed substance in moles. Finish by converting calculated moles to desired dimension.

Hints for remaining problems:

Divide given moles of reactant substances by respective coefficients, the smaller value is the limiting reactant. Work problem based on moles (not the divide value. That's just for ID of Limiting Reactant). All other reactants will be in excess.

for problem 5 ...

Given       8S(l)        +         4Cl₂(g) => 4S₂Cl₂(l)

Given:  0.433mole       0.500mole

LR        0.433/8             0.500/4

            = 0.054             = 0.125

Limiting Reactant => Sulfur

Work problem from given 0.433 mole sulfur. Cl₂ will be in excess on completion of rxn.

Summary:

- convert data to moles

- divide mole values calculated by respective coefficient => smaller value is limiting reactant.

- use mole ratios to determine results, NOT the divided by value <=> this is only for ID of limiting reactant.

If ya need more, put question in comments. I'll get it. Now, If you do need additional input, before I do I will ask if you followed the hint suggestions, and your calculation results. Good luck :-)