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Scorpion4ik [409]
3 years ago
15

If you start with 30.0 grams of sodium fluoride, how many grams of magnesium fluoride will be produced?

Chemistry
1 answer:
Cloud [144]3 years ago
7 0

Answer:

22.2 g of MgF₂

Explanation:

We star from the reaction:

NaF → sodium fluoride

Mg → Magnesium

2NaF  +  Mg →  MgF₂  +  2Na

2 moles of sodium fluoride react to 1 mol of Mg in order to produce 1 mol of magnesium fluoride and 2 moles of sodium.

We convert mass of the reactant to moles:

30 g . 1mol / 41.98 g = 0.715 mol

We assume that Mg is in excess

As ratio is 2:1, if we have 0.715 moles of NaF we may produce the half of moles, of MgF₂

0.715 mol /2 = 0.357 moles of MgF₂

We convert moles to mass: 0.357 mol . 62.30g /mol = 22.2 g

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5 0
4 years ago
A 11.97g sample of NaBr contains 22.34 % Na by mass. Considering the law of constant composition (definite proportions), how man
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<h3>Answer:</h3>

2.125 g

<h3>Explanation:</h3>

We have;

  • Mass of NaBr sample is 11.97 g
  • % composition by mass of Na in the sample is 22.34%

We are required to determine the mass of 9.51 g of a NaBr sample.

  • Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.

In this case,

  • A sample of 11.97 g of NaBr contains 22.34% of Na by mass
  • Therefore;

A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass

  • But;

% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100

  • Therefore;

Mass of the element = (% composition of an element × mass of the compound) ÷ 100

Therefore;

Mass of sodium = (22.34% × 9.51 g) ÷ 100

                         = 2.125 g

Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g

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3 years ago
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