I need help please if you only you hear my voice youd know how incredibly desperate I am please help me with my Chemistry homework as soon as possible please

Question

1.) S   + 6 HNO3 --> H2SO4 + 6 NO2 + 2 H2O

In the above equation how many moles of water can be made when 163.6 grams of HNO3 are consumed?

Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element	Molar Mass
Hydrogen	1
Nitrogen	14
Sulfur	32
Oxygen	16

2.) 3 Cu + 8HNO3 --> 3 Cu(NO3)2   + 2 NO + 4 H2O

In the above equation how many moles of water can be made when 118.6 grams of HNO3 are consumed?

Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0

Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:

Element	Molar Mass
Hydrogen	1
Nitrogen	14
Copper	63.5
Oxygen	16

please helps please

Answer 1

1) For the first problem, you start by looking at HNO3 and because they tell you that HNO3 is consumed, you know that this is leading the reaction.

Take HNO3 convert the grams you have been given to moles and then you can follow the stoichiometry to know what will be yielded.

to convert to moles use dimensional analysis and the atomic weights or masses you have been given here in the problem.

1 H = 1

1 N= 14

3 O = 16 x3= 48

atomic. weight for HNO3 = 63g/mol

using these units convert to moles

1 63g

___________ X ________

163.6g mol

grams cancel out and you have 0.8351 mol-1, but you will have to take the inverse to get this to just moles. so by doing 1/0.8351 this will give you the exact moles of 1.197 moles.

Then look at equation given in problem, is the equation balanced?

S   + 6 HNO3 --> H2SO4 + 6 NO2 + 2 H2O

You have:

6 Hydrogens on both sides

6 Nitrogens on both sides

18 Oxygens on both sides

1 Sulfur on both sides

you have a balanced equation

So go according to the balanced equation use the coeffecients in front of each species on the reactant and product side:

If 6 moles HNO3 ------> 2 moles H20,

then 1.197 moles HNO3 -----> X moles H20

cross multiply and divide,

6X = 2(1.197)

X = 2(1.197)

------------

6

X = 2.394

-------

6

X= 0.399 = 0.4 moles of water, H2O when rounding to the nearest tenths place

So you will get 0.4 moles of water, H2O from 1.197 moles of HNO3, which you received from 163.6g of the molecule/acid HNO3.

2) You use the same concept and process for this problem as well:

convert your grams to moles of the nitric acid HNO3 here we have it at 118.6 g of HNO3

the atomic weight of HNO3 is still: 63g see above for how it was derived

now taking the 118.6 g

1/118.6g x 63g/mol = 0.53119 mol-1, so you still have to take the inverse because moles is on bottom,

so 1/0.53119 = 1.8825 moles of HNO3

Now using this balanced equation, find out how many moles of water H2O can be yielded:

3 Cu + 8HNO3 --> 3 Cu(NO3)2   + 2 NO + 4 H2O

If 8 moles of HNO3 -------> 4 moles of H2O

then 1.8825 moles HNO3------> X moles of H2O

Cross multiply divide and solve for X

8X= 1.8825(4)

X= 1.8825(4)

---------

8

X= 7.53

--------

8

X = 0.94125 = 1.0 moles when rounded to the tenths place. This is your moles of water produced or yielded from this reaction with your HNO3 being 1.8825 moles and 118.6 g of the HNO3.