Chemistry Help! Need help with atleast one.

Question

Consider the reaction K2S(aq) + Co(NO3)2(aq) --> 2KNO3(aq) + CoS(s)

What mass of CoS is produced from 175 mL of 0.115 M Co(NO3)2? _____ g

 

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Consider the reaction K2S(aq) + Co(NO3)2(aq) --> 2KNO3(aq) + CoS(s)

What mass of CoS is produced from 4.84 g Co(NO3)2 and 115 mL of 0.225 M K2S solution? _____ g

Answer 1

Editing Frank's answer to clarify some things...

 

Gram Atomic weights (can get more digits from the periodic table, if your teacher is particular about significant digits, you should use more precise numbers)
K - 39
S - 32
Co - 59
N - 14
O - 16

 

Formula Weights for the relevant substances in grams per mole (g/mol)

You might want to re-calculate these with more digits/significant figures. 

If you are doing this for a multiple choice test, it probably won't matter.

If you are doing online homework or if your teacher is particular about sig figs, it might.

K2S - 39*2 + 32 = 110 g/mol
Co(NO3)2 - 59 + (14+16*3) * 2 = 183  g/mol
KNO3 - 39 + 14+ 16 * 3 = 101 g/mol
CoS - 59 + 32 = 91 g/mol

 

Molarity (M) is moles of solute per liter of solution, which can also be expressed as moles of solute per 1000 mL. If you have "moles" the abbreviation is "mol", not M. 

a) 175 ml of 0.115M Co(NO3)2 contains

175 mL x (0.115 mol Co(NO3)2 / 1000 mL) = 0.020125 mol Co(NO3)2

From the balanced chemical equation, we know that 1 mol Co(NO3)2 produces 1 mol CoS

We should be able to produce 0.020125 mol CoS x (91 g CoS / 1 mol CoS) = 1.831375 g CoS

Should really end up with 3 sig figs so 1.83 g CoS

But since we didn't use all of the sig figs in the atomic and formula weights, we might be off a little in the last digit.  However, this should be close.


b) 4.84 g Co(NO3)2 x (1 mol Co(NO3)2 / 183 g) = 0.026448 mol Co(NO3)2
115 ml of 0.225M K2S = 0.025875 mol K2S

 

Each of these reacts in a 1:1 mole ratio to form the product CoS, so if we use up all of the reactant, we should be able to produce that many moles of CoS.  However, one reactant often runs out first and limits the reaction, in this case, K2S. So we should only be able to make 0.025875 mol of CoS product.

Take the smallest (because that is all you can make!)
0.025875 mol CoS x (91 g CoS / 1 mol CoS) = 2.354625 g CoS formed

Again, we should only have 3 sig figs here based on the original data

2.35 g CoS should be formed (but might be off due to not using all of the digits in the beginning)

 

Hope the added detail helps, please note I did not re-calculate anything, just added descriptions and hints.  ;)

Answer 2

Gram Molecular weights
K - 39
S - 32
Co - 59
N - 14
O - 16

K2S - 39*2 + 32 = 110
Co(NO3)2 - 59 + (14+16*3) * 2 = 183
KNO3 - 39 + 14+ 16 * 3 = 101
CoS - 59 + 32 = 91

a) 175 ml of 0.115M = 0.020125M
0.20125M * 91 g = 1.831375 g


b) 4.84 g / 183 g = 0.026448M
115 ml of 0.225M = 0.025875M

 

Take the smallest
0.025875M * 91 g = 2.354625 g