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Masteriza [31]
3 years ago
10

How many moles of water molecules are found in 3.80 moles of zinc(ii) sulfate heptahydrate?

Chemistry
1 answer:
Sindrei [870]3 years ago
3 0
Zinc (ii) sulfate heptahydrate has a formula of ZnSO4.7H2O 
Therefore, note that one mole of ZnSO4.7H2O, contains 7 moles of H2O,
Hence, 3.8 moles of ZnSO4.7H2O will contain;
   3.8 × 7/1
  = 26.6 moles of H2O
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Answer:

1027.62 g

Explanation:

For B_2H_6  :-

Mass of B_2H_6  = 296.1 g

Molar mass of B_2H_6  = 27.66 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{296.1\ g}{27.66\ g/mol}

Moles\ of\ B_2H_6= 10.705\ mol

From the balanced reaction:-

B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}

1 mole of B_2H_6 react with 3 moles of oxygen

Thus,

10.705 mole of B_2H_6 react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

<u>Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g</u>

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3 years ago
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Answer:

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