Question

12. How many grams of glucose (C6H1206) is produced if 17.3 mol of H20 is reacted according to this

Answer 1

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.