Finding mole ratios from chemical formulae

Chemistry

1 answer:

Levart [38]1 year ago

7 0

7 moles of oxygen are in the sample.

According to the chemical formula, each mole of nickel tetracarbonyl contains 4 moles of C atoms. Simply convert it into a fraction by putting the original solution in the denominator and the diluted solution in the numerator if you need to determine the concentration ratio between two solutions. The V/n ratio for each gas must be the same if the two gases are at the same temperature and pressure. The volume ratio of two gases at the same temperature and pressure is equal to their molar ratio. The mole ratio of C to O is 1 : 1

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How many moles of salt will you be using in this recipe?

user100 [1]

5 molessss booo booo

6 0

3 years ago

Ammonium dihydrogen phosphate, formed from the reaction of phosphoric acid with ammonia, is used as a crop fertilizer as well as

Tema [17]

Answer:

a)The mass percentages of nitrogen and phosphorus in the compound is 24.34% and 26.95% respectively.

b) 14.78 grams ammonia is incorporated into 100. g of compound.

Explanation:

a) Ammonium dihydrogen phosphate that is $NH_6PO_4$ .

Molecular mass of ammonium dihydrogen phosphate = M

M = 115 g/mol

Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

Percentage of nitrogen:

$\frac{1\times 28 g/mol}{115 g/mol}\times 100=24.34\%$

Percentage of phosphorus:

$\frac{1\times 31g/mol}{115 g/mol}\times 100=26.95\%$

b) Percentage of ammonia in 1 molecule of ammonium dihydrogen phosphate.

Molar mass of ammonia = 17 g/mol

$\%=\frac{17 g/mol}{115 g/mol}\times 100=14.78\%$

Amount of ammonia in 100 grams of compound:

14.78% of 100 g of ammonium dihydrogen phosphate:

$\frac{14.78}{100}\times 100 g=14.78 g$

14.78 grams ammonia is incorporated into 100. g of compound.

4 0

3 years ago

Ammonia (NH3) reacts with sulfuric acid to form ammonium sulfate. How many grams of ammonium sulfate are obtained from 8.73 g of

Gwar [14]

I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra