How many moles of sodium chloride can be produced if 29g of Magnesium Chloride are reacted in the reaction 2NaNO3 + MgCl> 2NaCl+Mg (NO3)2?
Did you try doing the Stoichiometric equation?
Mg - 24.31 g
Cl - 35.45 g
Total Mass - 59.76 g
29g MgCl (1 mol MgCl/59.76g)= the answer.
.62 mole
To find the number of moles of sodium chloride (NaCl) produced, we need to use the given mass of magnesium chloride (MgCl2) and the balanced chemical equation.
The balanced equation is: 2NaNO3 + MgCl2 → 2NaCl + Mg(NO3)2
First, find the molar mass of MgCl2. The molar masses are:
Molar mass of Mg = 24.31 g/mol
Molar mass of Cl = 35.45 g/mol
Molar mass of MgCl2 = 24.31 g/mol + (2 × 35.45 g/mol) = 95.21 g/mol
Now, we can use the given mass of MgCl2 (29g) and the molar mass to calculate the number of moles.
Number of moles of MgCl2 = Given mass / Molar mass
Number of moles of MgCl2 = 29g / 95.21 g/mol ≈ 0.305 moles
Since the balanced equation shows that 1 mole of MgCl2 produces 2 moles of NaCl, we can now determine the number of moles of NaCl produced.
Number of moles of NaCl = 2 × Number of moles of MgCl2
Number of moles of NaCl = 2 × 0.305 moles ≈ 0.61 moles
Therefore, approximately 0.61 moles of sodium chloride can be produced if 29g of magnesium chloride is reacted.