1+7+(8x3)
=32
=32x2 (2 moles)
=64
4 moles HNO3 (63.018 grams/1 mole HNO3) = 252 grams nitric acid ================
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will get half the number of moles of NO2. So, you will have 0.2 moles of nitric acid.
20 moles of NaOH needed to neutralize 20 moles of nitric acid
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will end up with half the moles of nitrogen dioxide (NO2)...so you will have 0.2 moles.
63 if it is anhydrous
4 moles HNO3 (63.018 grams/1 mole HNO3) = 252 grams nitric acid ================
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will get half the number of moles of NO2. So, you will have 0.2 moles of nitric acid.
20 moles of NaOH needed to neutralize 20 moles of nitric acid
Starting with the formula: 2HNO3 --> H2O + NO2 If you have 0.4 moles of nitric acid (HNO3), you will end up with half the moles of nitrogen dioxide (NO2)...so you will have 0.2 moles.
63 if it is anhydrous
The answer is 0,07 moles Ca(OH)2.
0.1538 grams
3Cu + 8HNO3 ----> 3Cu(NO3)2 + 2NO + 4H2O is the equation if it is dilute nitric acid. In concentrated nitric acid the equation is different. So 3 moles of copper produce 2 moles of NO. Therefore it requires 6 moles of copper to produce 4 moles of NO.
I think you meant " How many moles of acetic acid in 25 grams of acetic acid? " We will use the chemist formula for acetic acid, 25 grams C2H4O2 (1 mole C2H4O2/60.052 grams) = 0.42 mole acetic acid =================
7
The mass of sulfuric acid is 490,395 grams.
How many moles of H+ ions are present in the following aqueous solution? 1,410 mL of 0.32 M nitric acid