Answer:
O2 is limiting reactant
Explanation:
To find the limiting reactant we need to convert the mass of each reactant to the moles using the formula weight. And, as 1 mole of C6H12O6 reacts with 6 moles of O2, we can know wich reactant will be over first (Limiting reactant) as follows:
<em>Moles C6H12O6:</em>
650g * (1mol/180.16g) = 3.608 moles C6H12O6
<em>Moles O2:</em>
650g * (1mol/32g) = 20.31 moles O2
Now, for a complete reaction of 3.608 moles of C6H12O6 are required:
3.608 moles C6H12O6 * (6mol O2 / 1mol C6H12O6) = 21.65 moles O2
As there are just 20.31 moles of O2,
<h3>O2 is limiting reactant</h3>
Metals are insulators and malleable. Hope this helps even though I'm late :)
O magnesium Mg is the answer
Use formula: Initial Pressure x Initial Volume/Initial temperature = Final pressure x Final Volume/Final Temperature => 17.15L
34g C * ( 1 mol / 12.0107 ) * ( 1 mol H2 / 1 mol C ) * ( <span>2.01588 g / 1 mol H2 ) = 5.70657164028741 g H2 = 5.7 g H2
Convert grams of C to moles of C using the given amount of grams and the molar mass ( 12.0107 g/mol ).
Gather the mole ratio from the coefficients in the balanced equation and multiply by the ratio.
Convert moles of H2 to grams of H2 </span> using the given amount of grams and the molar mass ( 2.01588 g/mol )<span>.
Revise your answer to have the correct number of significant figures. </span>
