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professor190 [17]
3 years ago
7

A hydrate of magnesium sulfate has a mass of 13.52 g. The sample is heated until no water remains. The anhydrate has a mass of 6

.60 g. Find the formula and name of the hydrate.
Chemistry
1 answer:
choli [55]3 years ago
4 0

Answer:

molecular formula = MgSO4.7H2O

Name is Magnesium sulfate heptahydrate

Explanation:

A hydrate of magnesium sulfate has a mass of 13.52 g. This sample is heated until no water remains. The MgSO4 anhydrate has a mass of 6.60 g. Find the formula and name of the hydrate.

given

mass 1 = 13.52 g

mass 2 = 6.60 g

solution

MgSO4 × H2O

first we ge here mass of hydrate part that is here

mass of hydrate part = (13.52 g - 6.60 g)

mass of hydrate part = 6.92 g

and

now we get here percentage of hydrate that is

percentage of hydrate = (6.92 / 13.52) x 100

percentage of hydrate = 51.18 %

so molar mass will be

molar mass of hydrate MgSO4 = 120.366 g/mol = 49.82 %

and for 100% = 241.6 g

so

mass of H2O = 121.23 g

and

moles of H2O = 7

so we can say that molecular formula will be MgSO4.7H2O

and name is Magnesium sulfate heptahydrate

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If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc
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To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

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Thus 69.72 kg of cryolite will be produced.

8 0
3 years ago
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