Calculate the mass of h2o produced by metabolism of 3.0 kg of fat, assuming the fat consists entirely of tristearin (c57h110o6), a typical animal fat, and assuming that during metabolism, tristearin reacts with o2 to form only co2 and h2o.

You have to start by balancing the equation

This is the equation balanced:



Relation 2 mols of triestarin: 110 mol of water

To pass 2 mols of triesetarin you have to multiply by the molar mass of the compound.

The molar mass of the triestarin is calculated as 57*atomic mass of C + 110 atomic mass of H + 6*atomic mass of oxygen = 57*12 + 110*1 + 6*16 = 890 g/mol

Then 2 mols = 2 * 890 g/mol = 1780 g

Now pass 110 mol of water to grams using the molar mass of water (18 g/mol)

110 mol * 18 g/mol = 1980 g of water

Now the relation in grams is 1780 g of tristearin / 1980 g of water = 3000 g of triestarin / x grams of water.

Solving for x, x = 3000 g of triestarin * 1980 g of water /1780 g of triestarin

x = 3337 g of water = 3.3 kg of water

3.3 kg of water

Ch4 and ph3 are both covalent since they are composed both of non-metals.    ch4 is non- polar because of  london forces and has a geometry of tetrahedral  ph3 , on the other hand, is umbrella shaped polar and has dipole-dipole intermolecular force (imf) . since ph3 is polar, it has a stronger imf