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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