The tungsten metal used for filaments in light bulbs is made by reaction of tungsten trioxide with hydrogen: wo3(s)+3h2(g)→w(s)+3h2o(g)
part a how many grams of tungsten trioxide must you start with to prepare 1.80 g of tungsten? (for wo3, mw = 231.8 amu.)
part b how many grams of hydrogen must you start with to prepare 1.80 g of tungsten?

A) 2.27 grams of tungsten trioxide must be needed to prepare 1.80 g of tungsten.

B)To prepare 1.80 g of tungsten we will need 0.0587 grams of hydrogen gas.

Explanation:

A) Mass of tungsten prepared = 1.80 g

Moles of tungsten =

According to reaction, 1 mol of tungsten is obtained from 1 mole of tungsten trioxide.

Then 0.009783 moles of tungsten will be obtained from:

of tungsten trioxide

Mass of 0.009783 moles of tungsten trioxide :

0.009783 mol × 231.8 g/mol = 2.27 g

2.27 grams of tungsten trioxide must be needed to prepare 1.80 g of tungsten.

B) According to reaction,1 mol of tungsten is produced by 3 moles of hydrogen gas

Then 0.009783 moles of tungsten are produced by :

Mass of 0.02935 moles of hydrogen gas:

0.02935 mol × 2 g/mol =0.0587 g

To prepare 1.80 g of tungsten we will need 0.0587 grams of hydrogen gas.

it's filtration, but this is an assumption so my apologies if i got this