First you have to get to a balanced equation
2KNO3+1Mg=1Mg(NO3)2+2K
Then you do a mole to mole equation
2.5molKNO3x1mol Mg(NO3)2/(over)2molKNO3=1.25mol Mg(NO3)2
So your answer is 1.25mol Mg(NO3)2
But if you follow the rules of signicicant figures you need to have as many digits in your answer as the given. 2.5mol has 2 significant figures(digits) so your answer needs to have 2 digits
Rounding to two digits we get 1.3molMg(NO3)2
Your answer is 1.3molMg(NO3)2
No, the molecular weight of KNO3 (Potassium Nitrate) cannot be 4.2. The actual molecular weight of KNO3 is approximately 101.1 g/mol, derived from the atomic masses of potassium (39.1 g/mol), nitrogen (14.0 g/mol), and oxygen (16.0 g/mol) in the compound.
Get moles by; Molarity = moles of solute/Liters of solution 0.50 M KNO3 = moles KNO3/2.0 L = 1.0 mole KNO3 Now find grams of 1.0 mole KNO3 1.0 mole KNO3 (101.11 grams/1 mole KNO3) = 101.11 grams KNO3 needed call it 100 grams
Divide 175 by 750 and multiply by 1000 to get g/l. This gives you 233.3333g/litre. The molecular weight for KNO3 is 101.1g/mol. 233.333/101.1 is 2.31 molar.
Molality = moles of solute / kilograms of solventm = 5.10 / 4.47m = 1.14 molal
1 mol K X (39.10 g K / mol K) = 39.10 g K1 mol N X (14.01 g N / mol N) = 14.01 g N3 mol O X (16.00 g O / mol O) = 48.00 g OMolar mass of KNO3 = 101.11 g/mol
The molecular weight of potassium nitrate (KNO3) is 101.1 grams/mole.K = 39.1 + N = 14.0 + O = 3*16.0 = 101.1Now divide 6.5 by 101.1 to calculate # of moles => 0.064 moles KNO3
No, the molecular weight of KNO3 (Potassium Nitrate) cannot be 4.2. The actual molecular weight of KNO3 is approximately 101.1 g/mol, derived from the atomic masses of potassium (39.1 g/mol), nitrogen (14.0 g/mol), and oxygen (16.0 g/mol) in the compound.
Get moles by; Molarity = moles of solute/Liters of solution 0.50 M KNO3 = moles KNO3/2.0 L = 1.0 mole KNO3 Now find grams of 1.0 mole KNO3 1.0 mole KNO3 (101.11 grams/1 mole KNO3) = 101.11 grams KNO3 needed call it 100 grams
How many moles of CO2 are produced when 2.1 mol of C2H2 react?
The molarity of the KNO3 solution is 0.544 M. This is calculated by dividing the moles of KNO3 (1.1 mol) by the total solution volume in liters.
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)
Divide 175 by 750 and multiply by 1000 to get g/l. This gives you 233.3333g/litre. The molecular weight for KNO3 is 101.1g/mol. 233.333/101.1 is 2.31 molar.
Molality = moles of solute / kilograms of solventm = 5.10 / 4.47m = 1.14 molal
CaC12
1 mol K X (39.10 g K / mol K) = 39.10 g K1 mol N X (14.01 g N / mol N) = 14.01 g N3 mol O X (16.00 g O / mol O) = 48.00 g OMolar mass of KNO3 = 101.11 g/mol
To find the total moles of KNO3 needed, use the formula: moles = molarity x volume (in liters). So, moles = 2.0 mol/L x 1.5 L = 3.0 moles of KNO3. Therefore, 3.0 moles of KNO3 need to be dissolved in water to make 1.5 liters of a 2.0 M solution.
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 → 2NH3 From the equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, when 1.2 moles of H2 react, we can calculate the moles of NH3 produced as: 1.2 mol H2 * (2 mol NH3 / 3 mol H2) = 0.8 mol NH3.