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3 years ago

30.5 g of sodium metal reacts with a solution of excess lithium bromide. How many grams

Chemistry

1 answer:

vagabundo [1.1K]3 years ago

6 0

Answer:

9.28 g of Lithium metal

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

Na + LiBr —> NaBr + Li

Next, we shall determine the mass of sodium metal, Na that reacted and the mass of lithium metal, Li produced from the balanced equation.

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 1 x 23 = 23 g

Molar mass of Li = 7 g/mol

Mass of Li from the balanced equation = 1 x 7 = 7 g

Summary:

From the balanced equation above,

23 g of Na reacted to produce 7 g of Li.

Now, we can obtain the mass of lithium metal, Li produced by reacting 30.5 g of sodium metal, Na as shown:

From the balanced equation above,

23 g of Na reacted to produce 7 g of Li.

Therefore, 30.5 g of Na will react to produce = (30.5 x 7)/23 = 9.28 g of Li.

Therefore, 9.28 g of Lithium metal, Li were produced from the reaction.

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Answer:

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3 years ago

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3 years ago

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Tresset [83]

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3 years ago

Not sure how to do this

faust18 [17]

Answer:

0.0991 M

Explanation:

Step 1: Write the neutralization reaction between oxalic acid and sodium hydroxide.

H₂C₂O₄ + 2 NaOH = Na₂C₂O₄ + 2 H₂O

Step 2: Calculate the moles of oxalic acid

The molar mass of H₂C₂O₄ is 90.03 g/mol. The moles corresponding to 153 mg (0.153 g) are:

$0.153g \times \frac{1mol}{90.03g} = 1.70 \times 10^{-3} mol$

Step 3: Calculate the moles of sodium hydroxide

The molar ratio of H₂C₂O₄ to NaOH is 1:2.

$1.70 \times 10^{-3} molH_2C_2O_4 \times \frac{2molNaOH}{1molH_2C_2O_4} = 3.40 \times 10^{-3} molNaOH$

Step 4: Calculate the molarity of sodium hydroxide

$\frac{3.40 \times 10^{-3} mol}{34.3 \times 10^{-3} L} = 0.0991 M$

4 0

4 years ago