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3 years ago

10. Copper(i) bromide reacts with magnesium metal: 2 CuBr + Mg → 2 Cu + MgBrz

Chemistry

2 answers:

velikii [3]3 years ago

5 0

Answer:

72.6 grams

Explanation:

I got this answer through stoichiometry. For every 1 mole of Mg, 2 moles of CuBr are consumed. Because of this, multiply the moles of Mg by 2. Then, convert moles to grams.

Inessa05 [86]3 years ago

3 0

Answer:

72.611g

Explanation:

Step1:

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

2CuBr + Mg → 2Cu + MgBr2

Step 2:

Let us calculate the mass of Mg in 0.253 mole of Mg. This is illustrated below:

Molar Mass of Mg = 24g/mol

Mole of Mg = 0.253 mole

Mass of Mg =?

Mass = number of mole x molar Mass

Mass of Mg = 0.253 x 24

Mass of Mg = 6.072g

Step 3:

Let us calculate the mass of CuBr and the mass of Mg that reacted from the balanced equation. This is illustrated below:

Molar Mass of CuBr = 63.5 + 80 = 143.5g/mol

Mass of CuBr from the balanced equation above = 2 x 143.5 = 287g

Molar Mass of Mg = 24g/mol

Step 4:

The mass of CuBr consumed by 0.253 mole ( i.e 6.072g) of Mg can be obtained as follow:

From the balanced equation:

2CuBr + Mg → 2Cu + MgBr2

287g of CuBr were consumed by 24g of Mg.

Therefore, Xg of CuBr will be consume by 6.072g of Mg i.e

Xg of CuBr = (287x6.072)/24

Xg of CuBr = 72.611g

Therefore, 72.611g of CuBr is consumed by 0.253 mole of magnesium

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