C + O2= CO2
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
There are 1.92 × 10^23 atoms Mo in the cylinder.
<em>Step 1</em>. Calculate the <em>mass of the cylinder
</em>
Mass = 22.0 mL × (8.20 g/1 mL) = 180.4 g
<em>Step 2</em>. Calculate the<em> mass of Mo
</em>
Mass of Mo = 180.4 g alloy × (17.0 g Mo/100 g alloy) = 30.67 g Mo
<em>Step 3</em>. Convert <em>grams of Mo</em> to <em>moles of Mo
</em>
Moles of Mo = 30.67 g Mo × (1 mol Mo/95.95 g Mo) = 0.3196 mol Mo
<em>Step 4</em>. Convert <em>moles of M</em>o to <em>atoms of Mo
</em>
Atoms of Mo = 0.3196 mol Mo × (6.022 × 10^2<em>3</em> atoms Mo)/(1 mol Mo)
= 1.92 × 10^23 atoms Mo
The correct answer is "B. Make a Hypothesis".
