When 1.00 g of boron is burned in o2 (g) to form b2o3 (s), enough heat is generated to raise the temperature of 733 g of water from 18.0 ∘c to 38.0 ∘c?

Question

Carina Lowe

Chemistry

15 July, 15:48

When 1.00 g of boron is burned in o2 (g) to form b2o3 (s), enough heat is generated to raise the temperature of 733 g of water from 18.0 ∘c to 38.0 ∘c?

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Puddles · Answer 1

Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J / (g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us. 0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3 ...0925/2 =.0462 moles ... so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.