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Angelina_Jolie [31]
3 years ago
6

A certain crunch cereal contains 11.0 grams of sugar(sucrose, C12H22O11) per serving size of 60.0 gram. How many

Chemistry
1 answer:
Paladinen [302]3 years ago
5 0

Answer:

51.859 grams

Explanation:

Given:

Mass of sugar in the crunch cereal per serving= 11.0 grams

Mass of cereal served per serving = 60.0 grams

Molar mass of the sugar (C₁₂H₂₂O₁₁) = ( 12 × 12 + 22 × 1 + 16 × 11 ) = 342 grams

Number of moles of sugar per serving = Mass / Molar mass

= 11.0 / 342

= 0.03216 moles

Now,

0.03216 moles sugar requires 60 grams cereal

for 0.0278 moles of sugar, cereal required = (60 / 0.03216) × 0.0278

or

for 0.0278 moles of sugar, cereal required = 51.859 grams

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<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

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Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol

For the given chemical reaction:

3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)

  • <u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from \frac{3}{3}\times 0.252=0.252mol of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g

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By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from \frac{2}{3}\times 0.252=0.168mol of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

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