Answer:
43.2 moles of carbon dioxide are required and 421g of glucose could be produced
Explanation:
Based on the reaction:
6CO2 + 6H2O → C6H12O6 + 6O2
1 mole of glucose, C6H12O6, requires 6 moles of carbon dioxide. 7.2moles of glucose requires:
7.2mol C6H12O6 * (6mol CO2 / 1mol C6H12O6) =
<h3>43.2 moles of carbon dioxide are required</h3><h3 />
618g of CO2 -Molar mass: 44.01g/mol- are:
618g * (1mol / 44.01g) = 14.04moles CO2
Moles C6H12O6:
14.04moles CO2 * (1mol C6H12O6 / 6mol CO2) = 2.34moles C6H12O6
Mass glucose -Molar mass: 180.156g/mol-
2.34moles C6H12O6 * (180.156g / mol) =
<h3>421g of glucose could be produced</h3>
Good questions,ideas,and observations
The right answer for the question that is being asked and shown above is that: "<span>b. number/timed." Reaction Rate refers to the </span> speed of reaction<span> for a reactant or product in a particular </span>reaction<span> is intuitively defined as how fast or slow a</span>re action<span> takes place.</span>
