Author Topic: Hmmmmm..... Help me guys??? (Read 1025 times) Tweet Share

zibb3r · « **on:** February 03, 2011, 07:12:22 pm »

I have 2 questions:

1) 16.0 grams of potassium hydroxide and 12.0g of nitric acid were reacted to form potassium nitrate and water. Write an equation and determine which reactant is the limiting reactant and which reactant is in excess.

2) When a certain non-metal whose formula is X₈ burns in air, XO₃ forms. Write a balanced equaion for this reaction. Iff 120.0 g of oxygen gas is consumed completely, along with 80.0 g of X₈ , identify element X.

Thanks heaps guys!!!!!!

I love you all!!!!

m@tty · « **Reply #1 on:** February 03, 2011, 07:25:39 pm »

$KOH(aq)+HNO_3(aq) \to H_2O(l)+KNO_3(aq)$

(not sure about the states - been so long since I've had to work with chem equations =/ )

Therefore 1:1 ratio

$n(KOH)=\frac{16.0}{39.0+17.0}=0.276 \text{ mol}$

$n(HNO_3)=\frac{12.0}{1+14+48}=0.190 \text{ mol}$

Therefore nitric acid is the limiting reactant.

m@tty · « **Reply #2 on:** February 03, 2011, 07:35:28 pm »

2) When a certain non-metal whose formula is X8 burns in air, XO3 forms. Write a balanced equaion for this reaction. Iff 120.0 g of oxygen gas is consumed completely, along with 80.0 g of X8 , identify element X.

$X_8 +12O_2 \to 8XO_3$

$m(O_2)=120.0\text{ g}$

$n(O_2)=\frac{120.0}{32.0}=3.75 \text{ mol}$

Ratio - O_2 : XO_3 = 3:2

Therefore $n(XO_3)=\frac{2}{3} \times 3.75 = 2.50$

$M(XO_3)=\frac{200.0}{2.50}=80.0 \text{ gmol}^{-1}$ (mass of $XO_3$ is 200g due to law of conservation of mass.)

Therefore $M(X)=80.0-48.0=32.0\text{ gmol}^{-1}$ (subtract 48.0g as that is the mass of three oxygens.)

Therefore element X is sulfur.

Dr.Lecter · « **Reply #3 on:** February 03, 2011, 07:38:38 pm »

Quote from: m@tty on February 03, 2011, 07:35:28 pm

2) When a certain non-metal whose formula is X8 burns in air, XO3 forms. Write a balanced equaion for this reaction. Iff 120.0 g of oxygen gas is consumed completely, along with 80.0 g of X8 , identify element X.

$X_8 +12O_2 \to 8XO_3$

$m(O_2)=120.0\text{ g}$

$n(O_2)=\frac{120.0}{32.0}=3.75 \text{ mol}$

Ratio - O_2 : XO_3 = 3:2

Therefore $n(XO_3)=\frac{2}{3} \times 3.75 = 2.50$

$M(XO_3)=\frac{200.0}{2.50}=80.0 \text{ gmol}^{-1}$ (mass of $XO_3$ is 200g due to law of conservation of mass.)

Therefore $M(X)=80.0-48.0=32.0\text{ gmol}^{-1}$ (subtract 48.0g as that is the mass of three oxygens.)

Therefore element X is sulfur.

What do you use to get that type of output, instead of using basic characters on a keyboard?

m@tty · « **Reply #4 on:** February 03, 2011, 07:41:07 pm »

LaTeX - there is a basic introduction if you follow the link.

It is basic programming, essentially.

zibb3r · « **Reply #5 on:** February 03, 2011, 07:46:08 pm »

WOW! m@tty THaNKYOU!!! btw is there a rule against posting too many questions?

?

m@tty · « **Reply #6 on:** February 03, 2011, 07:52:42 pm »

That's alright, zibb3r; I'm happy to help.

No there is no limit on questions. As long as you're learning from them.

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Author Topic: Hmmmmm..... Help me guys??? (Read 1025 times) Tweet Share

zibb3r

Hmmmmm..... Help me guys???

m@tty

Re: Hmmmmm..... Help me guys???

m@tty

Re: Hmmmmm..... Help me guys???

Dr.Lecter

Re: Hmmmmm..... Help me guys???

m@tty

Re: Hmmmmm..... Help me guys???

zibb3r

Re: Hmmmmm..... Help me guys???

m@tty

Re: Hmmmmm..... Help me guys???

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