Answer:
1. 17.3 MOLES OF WATER WILL PRODUCE 518.4 G OF GLUCOSE.
2. 87.4 G OF HNO3 WILL PRODUCE 7.86 G OF NITROGLYCERIN WHEN REACTED WITH EXCESS GLYCEROL.
Explanation:
1. How many grams of glucose is produced from 17.3 mole of water?
Equation:
6CO2 + 6H20 ------> C6H12O6 + 6O2
From the reaction, 6 moles of carbon dioxide reacts with 6 moles of water to produce 1 mole of glucose
So therefore,
6 moles of water = 1 mole of glucose
Since 17.3 mole of water was used, we can first calculate the number of moles of glucose produced:
Then, we have:
6 moles of water = 1 mole of glucose
17.3 moles of water = ( 17.3 * 1/ 6) moles of glucose
= 2.883 moles of glucose
So we say 17.3 moles of water produces 2.883 moles of glucose
At standard conditions, 1 mole of a substance is its molar mass
Molar mass of water = 18 g/mol
Molar mass of glucose = 180 g/mol
From the reaction:
17.3 moles of water produces 2.883 moles of glucose
17.3 * 18 g of water produces 2.833 * 180 g of glucose
= 518.94 g of glucose.
From 17.3 mole of water, 518.4 g of glucose will be produced.
2.
C3H5(OH)3 + 3HNO3 -------> C3H5(ONO2)3 + 3H20
3 moles of HNO3 reacts to produce 1 mole of nitroglycerin
Molar mass of HNO3 = ( 1 + 14 + 16*3) = 63 g/mol
Molar mass of nitroglycerin = ( 12 *3 + 1 *5 + 16*6 + 14*3) = 179 g/mol
3 moles of HNO3 = 1 mole of nitroglycerin
3 * 63 g of HNO3 = 179 g of nitroglycerin
if 87.4 g of HNO3 were to be reacted, we have:
189 g of HNO3 = 179 g of nitroglycerin
87.4 g of HNO3 = ( 87.4 * 179 / 189) of nitroglycerin
= 7.86 g of nitroglycerin
So therefore, 87.4 g of HNO3 will produce 7.86 g of nitroglycerin when reacted with excess glycerol.
