Answer: 37.5grams of Cu(NO3)2
Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2
<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>
<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>
<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>
<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>
Answer:
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1) Chemical equation
<span>2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l)
2) Stoichiometric ratios
2 mol NH4Cl(s) : 54.8 KJ
3) Convert 24.7 g of NH4Cl into number of moles, using the molar mass
molar mass of NH4Cl = 14 g/mol + 4*1 g/mol + 35.5 g/mol = 53.5 g/mol
number of moles = mass in grams / molar mass
number of moles = 24.7 g / 53.5 g/mol = 0.462 moles
4) Use proportions:
2 moles NH4Cl / 54.8 kJ = 0.462 moles / x
=> x = 0.462 moles * 54.8 kJ / 2 moles = 12.7 kJ
Answer: 12.7 kJ
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