Question

Write the equation $6{H_2}O + 6C{O_2} \to {C_6}{H_{12}}{O_6} + 6{O_2}$ . How many grams of water are needed to make the above amount of glucose if you start with $88.0$ grams of Carbon dioxide?

Answer 1

In order to answer this question, we have to calculate the amount of water and carbon dioxide required to form one mole of glucose. This can be done by using the formula of number of moles which is the ratio of weight and gram molecular weight.

Complete step-by-step answer:
We have an equation $6{H_2}O + 6C{O_2} \to {C_6}{H_{12}}{O_6} + 6{O_2}$. This equation is the reaction between water and carbon dioxide to form glucose and oxygen. We should first calculate the amount of water and carbon dioxide used to produce one mole of glucose. Six moles of water is used to produce one mole of glucose then the weight of water would be $6 \times 18 = 108$ grams.

Similarly, $6$ moles of carbon dioxide is used to produce one mole of glucose. Weight of carbon dioxide would be $6 \times 44 = 264$. The weight of glucose produced would be its gram molecular weight $180$ . It is given that $88$ grams of carbon dioxide is used. We have to calculate the weight of water required to produce glucose.$264$ grams of carbon dioxide requires $108$ grams of water, then $88$ grams of carbon dioxide would require X grams of water.The value of X is calculated by cross multiplication as $X = \dfrac{{108 \times 88}}{{264}} = 36$

Therefore the required weight of water is $64$

Note: It is to be noted that the weight of one mole of a compound is equal to its gram molecular weight. In the above question $6$ moles of water and $6$ moles of carbon dioxide are used to produce one mole of glucose and hence their gram molecular weights are multiplied by $6$ to get the resultant weight.