Question

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?

Answer 1

Answer:

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:

$WO_3(s)+3H_2(g)\rightarrow W(s)+3H_2O(g)$

A) Mass of tungsten prepared = 1.80 g

Moles of tungsten =$\frac{1.80 g}{184 g/mol}=0.009783 mol$

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:

$\frac{1}{1}\times 0.009783 mol=0.009783 mol$ 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 :

$\frac{3}{1}\times 0.009783 mol=0.02935 mol$

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.