12
65
10 moles of nitrogen dioxide are needed to react with 5,0 moles of water.
When 3 moles Cu react 3 moles of copper nitrate are obtained.
The nunber of moles of oxygen is 2,5.
The answer is 3 moles of Cu)NO3)2..
65
3 moles KOH to neutralize 3 moles HNO3 : 3 mol OH- will react with 3 mol H+
40 moles of LiOH
NaOH = 40 Mwt so 15/40 moles present. This requires 15/40 moles of HNO3 from the above equation. The HNO3 contains 2 moles in 1000 ml and so 1 mole in 500 ml and therefore 500 x 15/40 = 137.5 mls required
4 moles HNO3 (63.018 grams/1 mole HNO3) = 252 grams nitric acid ================
10
.44 moles BeCl2
For this you need the atomic (molecular) mass of HNO3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. HNO3= 63.0 grams2.00 moles HNO3 × (63.0 grams) = 126 grams HNO3
7.8
Balanced equation first. 2CH4O + 3O2 -> 2CO2 + 4H2O 23.5 moles methanol (3 moles O2/2 mole CH4O) = 35.3 moles oxygen needed --------------------------------------
Balanced Equation. NaOH + HNO3 >> NaNO3 + H2O Now, Molarity = moles of solute/liters of solution 0.800M HNO3 + mol/2.50L mol of HNO3 = 2 2mol HNO3 (1mol NaOH/1molHNO3 )(39.998g NaOH/1mol NaOH ) = 79.996 grams
10 moles of nitrogen dioxide are needed to react with 5,0 moles of water.