How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
0,044 moles of NH3 can be produced.
1,4 moles carbon monoxide are produced.
6,2 moles of silver
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
1058 grams How to find it; moles (n) = mass (m) / molar mass (M) Or rearranging, m = n*M Looking at the periodic table to find M for BaBr2, we get, M = 137.33 + 79.9*2 = 297.13 So, m = 3.56 * 297.13 = 1058 g
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
0,044 moles of NH3 can be produced.
1,4 moles carbon monoxide are produced.
6,2 moles of silver
How many moles of CO2 are produced when 2.1 mol of C2H2 react?
If 1 mole of carbon reacts, 1 mole of CO is produced according to the balanced chemical equation for the reaction. Therefore, if 1.4 moles of carbon react, 1.4 moles of CO will be produced.
The answer is 3 moles of Cu)NO3)2..
The balanced chemical equation for the reaction between ammonia (NH3) and water (H2O) is: 4NH3 + 5O2 → 4NO + 6H2O. This means that for every 4 moles of ammonia, 6 moles of water are produced. Therefore, if 2 moles of ammonia are used, 3 moles of water vapor can be produced.