What are the coefficients when the equation below is balanced?

Chemistry

1 answer:

Sati [7]2 years ago

3 0

Answer:

1, 2, 2, 1

Explanation:

This is basically the equation we're trying to balance:

Cu + AgNO₃ → Ag + Cu(NO₃)₂

We start with Copper (Cu): There is 1 atom on the reactants'(left) side and 1 atom on the products'(right) side which means it's balanced

The second is Silver (Ag): There is 1 atom on the reactants' side and 1 atom on the products' side which means it's balanced

The third is Nitrogen (N) and there is 1 atom on the reactants' side and 2 atoms on the products' side. We balance it by multiplying the reactants' side by 2

Cu + 2AgNO₃ → Ag + Cu(NO₃)₂

Silver is now imbalanced so we multiply the silver atoms on the product's side by 2

Cu + 2AgNO₃ → 2Ag + Cu(NO₃)₂

If you count them now, the atoms on each side are the same number

1 Copper atom on each side

2 Silver atoms on each side

2 Nitrogen atoms on each side

6 Oxygen atoms on each side (2 sets of 3 oxygen atoms)

1Cu + 2AgNO₃ → 2Ag + 1Cu(NO₃)₂

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3 years ago

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Aleks04 [339]

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$