Question

2agno3+na2cro4 ag2cro4+2nano3For the following chemical reaction, how many moles of silver chromate (Ag2CrO4) will be produced from 8 mol of silver nitrate (AgNO3)?

For the following chemical reaction, what mass of silver chromate (in grams) will be produced from 3.91 mol of silver nitrate?

mg+cu(no3)2 mg(no3)2+cu Assuming an efficiency of 28.30%, calculate the actual yield of magnesium nitrate formed from 114.1 g of magnesium and excess copper(II) nitrate.

Answer 1

2AgNO₃+Na2CrO₄ --------> Ag₂CrO₄+2NaNO₃

In the above reaction, two moles of Silvernitrate reacts with one mole of sodium chromate to give one mole of silver chromate and two moles of sodium nitrite. Hence, if we take eight moles of silvernitrate we will obtain four moles of Ag₂CrO₄. i.e. 8AgNO₃+4Na₂CrO₄ --------> 4Ag₂CrO₄+8NaNO₃.

Silver nitrate molecular mass = 169.87 g/mol; Silver chromate mol.mass = 331.73 g/mol

OK now 3.91 mol of silvernitrate will give 3.91/2 moles of silverchromate (remember 8 moles of silvernitrate will yield 4 moles of silverchromate). So the weight of 1.955 moles of silverchromate is 1.955x331.73 = 648.53 grams.

Magnesium atomic mass = 24.3; MgNO3 molar mass = 148.3

114.1 g of magnesium is 4.695 moles.

4.695 moles of magnesium should yield (theoretically) 4.695 moles of magnesium nitrate as per the equation. 4.695 moles of MgNO3 is equivalent to 4.695x148.3 = 696 grams. This is 100% yield. So 28.30% yield would be equal to 196.968 grams.

Note: learn to calculate the number of moles from the given weight or the weight from the given moles. Then calculate the %yield.