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vaieri [72.5K]
4 years ago
7

How much NaAlO2 (sodium aluminate) is required to produce 2.15 kg of Na3AlF6?

Chemistry
1 answer:
Harrizon [31]4 years ago
5 0

Answer:

839 g

Explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

M_r:              81.97                209.94

                   NaAlO₂ + … ⟶ Na₃AlF₆ + …

Mass/g:                                   2150

1. Calculate the <em>moles of Na₃AlF₆ </em>

Moles of Na₃AlF₆ = 2150 × 1/209.94       Do the operation

Moles of Na₃AlF₆ = 10.24 mol Na₃AlF₆

2. Calculate the <em>moles of NaAlO</em>₂

The molar ratio is 1 mol NaAlO₂:1 mol Na₃AlF₆

Moles of NaAlO₂ = 10.24 × 1/1 = 10.24 mol NaAlO₂

3. Calculate the <em>mass of NaAlO₂ </em>

Mass of NaAlO₂ = 10.24 × 81.97             Do the multiplication

Mass of NaAlO₂ = 839 g

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According to this question, a 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O.

This means that the mass of sulfur in the 10.4g of CaSO4 is 10.4g - (3.06g + 4.89g) = 10.4g - 7.95g = 2.45g

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