What mass of HNO3 containing 10% W/W of acid will be required to dissolve 2.5g of CaCO3
First we write the balanced equation for the reaction as;
2HNO3 + CaCO3 "\\to" Ca (NO3)2 + H2O + CO2
Moles of CaCO3 in 2.5g "=" "\\dfrac{mass}{Molar mass}=\\dfrac{2.5g}{100.0869g\/mol}=0.024978mol"
Moles of HNO3 required;
Mole ratios of HNO3:CaCO3 = 2:1
Moles of HNO3 "=\\dfrac{2}{1}xmoles of CaCO3"
"=\\dfrac{2}{1}x0.024978 = 0.049957mol"
Concentration of acid = 10% w/w
That means 10g of HNO3 in 100g of solution
That is "\\dfrac{10g}{63.01g\/mo} moles" of HNO3 in 100g of solution
= 0.1587 moles of HNO3 in 100g of solution
Now, If 0.1587 moles are in 100g of solution;
0.049957 moles will be in?
"=\\dfrac{0.049957mol x 100g}{0.1587mol}"
"=" 31.479 g of 10% w/w of HNO3
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