# QuestionWatch

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#1
Got no idea how to do It, really appreciate some help. , answer is C btw
Q1) Calcium carbonate reacts with dilute nitric acid as follows:
CaCO3(s) + 2HNO3(aq) ----> Ca(NO3)2(aq) + H2O(l) + CO2(g)

0.05 mol of calcium carbonate was added to a solution containing 0.08 mol of
nitric acid.

Which of the following statements is true?
A 0.05 mol of carbon dioxide is produced.
B 0.08 mol of calcium nitrate is produced.
C Calcium carbonate is in excess by 0.01 mol.
D Nitric acid is in excess by 0.03 mol

Q2) If the price of one tonne (1000 kg) of sulfur, S, is £160, what is the cost (to the nearest pound) of the sulfur needed to make one tonne of sulfuric acid, H2SO4?
A £52
B £98
C £160
D £490

I did some simple maths calculation and got a number around 490, so chose D, but the answer is A
0
4 years ago
#2
(Original post by qatarownz)
Got no idea how to do It, really appreciate some help. , answer is C btw
Q1) Calcium carbonate reacts with dilute nitric acid as follows:
CaCO3(s) + 2HNO3(aq) ----> Ca(NO3)2(aq) + H2O(l) + CO2(g)

0.05 mol of calcium carbonate was added to a solution containing 0.08 mol of
nitric acid.

Which of the following statements is true?
A 0.05 mol of carbon dioxide is produced.
B 0.08 mol of calcium nitrate is produced.
C Calcium carbonate is in excess by 0.01 mol.
D Nitric acid is in excess by 0.03 mol

Q2) If the price of one tonne (1000 kg) of sulfur, S, is £160, what is the cost (to the nearest pound) of the sulfur needed to make one tonne of sulfuric acid, H2SO4?
A £52
B £98
C £160
D £490

I did some simple maths calculation and got a number around 490, so chose D, but the answer is A
First one is really simple. Look at the stoichiometry of the equation.

1:2 for the reactants.

So 0. 08 mol of hno3 should react with 0.04 mol of caco3 but there are 0.05 moles so its C.

Doing 2 now
0
4 years ago
#3
(Original post by qatarownz)
Got no idea how to do It, really appreciate some help. , answer is C btw
Q1) Calcium carbonate reacts with dilute nitric acid as follows:
CaCO3(s) + 2HNO3(aq) ----> Ca(NO3)2(aq) + H2O(l) + CO2(g)

0.05 mol of calcium carbonate was added to a solution containing 0.08 mol of
nitric acid.

Which of the following statements is true?
A 0.05 mol of carbon dioxide is produced.
B 0.08 mol of calcium nitrate is produced.
C Calcium carbonate is in excess by 0.01 mol.
D Nitric acid is in excess by 0.03 mol

Q2) If the price of one tonne (1000 kg) of sulfur, S, is £160, what is the cost (to the nearest pound) of the sulfur needed to make one tonne of sulfuric acid, H2SO4?
A £52
B £98
C £160
D £490

I did some simple maths calculation and got a number around 490, so chose D, but the answer is A
For question 2, I'm not sure what equation they use because with the contact process the answer is 52/2. So I'll make the assumption that they assume that the number of moles of sulfur is equal to sulfuric acid o, Mr of sulfuric acid equals 98 (16+16+16+16+32+1+1). Number of moles equal mass over Mr so 1/98 equals number of moles. 1/98 multiplied by 32 is mass of sulfur needed. That number multiplied by 160 is the answer
2
#4
(Original post by PharaohFromSpace)
For question 2, I'm not sure what equation they use because with the contact process the answer is 52/2. So I'll make the assumption that they assume that the number of moles of sulfur is equal to sulfuric acid o, Mr of sulfuric acid equals 98 (16+16+16+16+32+1+1). Number of moles equal mass over Mr so 1/98 equals number of moles. 1/98 multiplied by 32 is mass of sulfur needed. That number multiplied by 160 is the answer
Thanks a lot for the help, really appreciate it
0
4 years ago
#5
(Original post by qatarownz)
Thanks a lot for the help, really appreciate it
No problem 0
2 years ago
#6
pure s = 160\$
so now u calculate molecular mass of H2SO4
(1x2) 32 (16x4)
thats 98
then u find the percentage of S 32/98 x 100
its 32.65%
then u find 32.65% of 160\$
thats 52.24 = 52\$
1
2 years ago
#7
(Original post by Mịřäžūľ Īšľâm)
pure s = 160\$
Before helping someone, first check the date of the post you're replying to. If, like in this case, the post is a few years old, don't bother.

Also... welcome to TSR.
1
2 years ago
#8
Consider the following data:C(s) O2(g) o CO2(g) ǻH9 = –394 kJ mol–1Pb(s) ½O2(g) o PbO(s) ǻH9 = –217 kJ mol–1PbO(s) CO(g) o Pb(s) CO2(g) ǻH9 = –66 kJ mol–1 Calculate the value of the enthalpy change, in kJ mol–1, for the following reaction.C(s) ½O2(g) o CO(g)A –243B –111C 111D 243. .... any one pls help me how to solve this🙁
0
1 year ago
#9
(Original post by Mịřäžūľ Īšľâm)
pure s = 160\$
so now u calculate molecular mass of H2SO4
(1x2) 32 (16x4)
thats 98
then u find the percentage of S 32/98 x 100
its 32.65%
then u find 32.65% of 160\$
thats 52.24 = 52\$
Thanks for this!
0
1 year ago
#10
(Original post by sunjidaakter)
Consider the following data:C(s) O2(g) o CO2(g) ǻH9 = –394 kJ mol–1Pb(s) ½O2(g) o PbO(s) ǻH9 = –217 kJ mol–1PbO(s) CO(g) o Pb(s) CO2(g) ǻH9 = –66 kJ mol–1 Calculate the value of the enthalpy change, in kJ mol–1, for the following reaction.C(s) ½O2(g) o CO(g)A –243B –111C 111D 243. .... any one pls help me how to solve this🙁

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