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2 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

Chemistry

1 answer:

luda_lava [24]2 years ago

6 0

the balanced equation is CH4 + 2O2 -> CO2 + 2H20

moles = mass ÷ mr

moles of methane = 10 ÷ 16

= 10/16

= 5/8 moles

the moles of carbon dioxide are the same as methane because the ratio is 1:1

mass = moles × Mr

mass of carbon dioxide = 5/8 × 44

= 220 ÷ 8

= 27.5 grams

to find out whether it obeys the law of conservation of mass, you must find the mass of the reactants and the mass of the products. if they are the same, it obeys the law of conservation of mass. otherwise it doesnt.

the mass of oxygen is 32 × 5/8

= 160/8

= 20g

we already have the mass of methane (10g) so add it on

mass of reactants = 20 + 10 = 30g

find mass of water: 5/8 × 18

= 90/8

= 11.25g

mass of products = 11.25 + 27.5 = 38.75g

the mass of reactants vs products are not the same so it does not obey the law of conservation of mass

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AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

Answer: The theoretical yield of copper is 44.48 grams

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol$

The given chemical equation follows:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = $\frac{3}{2}\times 0.467=0.7005mol$ of copper

Now, calculating the mass of copper from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

$0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g$

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nordsb [41]

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