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6H2O + 6CO2 --> C6H12O6 + 6O2 How many grams of Glucose are made if you start with 88.0g of CO2?

1 Answer
Mar 14, 2018

Answer:

Approximately #60 \ "g of" \ C_6H_12O_6#.

Explanation:

We have the balanced equation (without state symbols):

#6H_2O+6CO_2->C_6H_12O_6+6O_2#

So, we would need six moles of carbon dioxide to fully produce one mole of glucose.

Here, we got #88 \ "g"# of carbon dioxide, and we need to convert it into moles.

Carbon dioxide has a molar mass of #44 \ "g/mol"#. So here, there exist

#(88color(red)cancelcolor(black)"g")/(44color(red)cancelcolor(black)"g""/mol")=2 \ "mol"#

Since there are two moles of #CO_2#, we can produce #2/6*1=1/3# moles of glucose #(C_6H_12O_6)#.

We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass.

Glucose has a molar mass of #180.156 \ "g/mol"#. So here, the mass of glucose produced is

#1/3color(red)cancelcolor(black)"mol"*(180.156 \ "g")/(color(red)cancelcolor(black)"mol")~~60 \ "g"# to the nearest whole number.

So, approximately #60# grams of glucose will be produced.