# 2agno3+na2cro4 ag2cro4+2nano3F

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

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

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 gof magnesium and excess copper(II) nitrate.

2AgNO3+Na2CrO4 ——–>Ag2CrO4+2NaNO3

In the above reaction, two moles of Silvernitrate reacts withone mole of sodium chromate to give one mole of silver chromate andtwo moles of sodium nitrite. Hence, if we take eight moles ofsilvernitrate we will obtain four moles ofAg2CrO4. i.e.8AgNO3+4Na2CrO4——–>4Ag2CrO4+8NaNO3.

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

OK now 3.91 mol of silvernitrate will give 3.91/2 moles ofsilverchromate (remember 8 moles of silvernitrate will yield 4moles of silverchromate). So the weight of 1.955 moles ofsilverchromate is 1.955×331.73 = 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) 4.695moles of magnesium nitrate as per the equation. 4.695 moles ofMgNO3 is equivalent to 4.695×148.3 = 696 grams. This is 100% yield.So 28.30% yield would be equal to 196.968grams.

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

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