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Helga [31]
3 years ago
9

How would i balance this equation?

Chemistry
1 answer:
Anna71 [15]3 years ago
6 0
Our coefficients would be:
C12H22O11: 1
O2: 12
CO2: 12
H2O: 11

Why?
With the coefficients, our reactants would be:
12 carbons, 22 hydrogens, and 35 oxygens.

To obey the law of conservation of mass, we would have to have an equal amount of each of those elements on the products side, since so much of that element needs to form an equal amount on the other side.

So, with those coefficiens, our products would be:
12 carbons total, 35 oxygens total, and 22 hydrogens total. If we compare to the reactants side, they are all equal!

And, since 1 or 11 cannot be simplifyed any smaller, we know the equation is completely balanced.

Hope I could help!
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a substance has a definite volume but no definite shape ? state whether this substance is a solid , a liquid , or a gas.​
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3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes
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Answer : The mass of MnCO_3 required are, 35 kg

Explanation :

First we have to calculate the mass of MnO_2.

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2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2

Molar mass of MnCO_3 = 115 g/mole

Molar mass of MnO_2 = 87 g/mole

Let the mass of MnCO_3 be, 'x' grams.

From the balanced reaction, we conclude that

As, (2\times 115)g of MnCO_3 react to give (2\times 87)g of MnO_2

So, xg of MnCO_3 react to give \frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg of MnO_2

And as we are given that the yield produced from the first step is, 65 % that means,

60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg

The mass of MnO_2 obtained = 0.4542x g

Now we have to calculate the mass of Mn.

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3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3

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Molar mass of Mn = 55 g/mole

From the balanced reaction, we conclude that

As, (3\times 87)g of MnO_2 react to give (3\times 55)g of Mn

So, 0.4542xg of MnO_2 react to give \frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg of Mn

And as we are given that the yield produced from the second step is, 80 % that means,

80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg

The mass of Mn obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g     (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of MnCO_3 required are, 35 kg

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